Free math worksheets

- A zero velocity vector requires both components to be zero simultaneously. - Examine whether either component can ever vanish. - One component may settle the question without solving the other equation.

1. A stopped particle must have both velocity components equal to \(0\). 2. The horizontal component \(t^2+1\) is positive for every real \(t\). 3. Therefore, the velocity vector can never be \(\langle0,0\rangle\), so the particle never stops.

The particle never stops.

- Velocity describes the instantaneous rate of change of position. - Treat the horizontal and vertical position components separately. - Evaluate the resulting rate vector at the stated time.

1. Differentiate the position components: \(\mathbf v(t)=\mathbf r'(t)=\langle2t,3\rangle\). 2. Evaluate at \(t=1\): \(\mathbf v(1)=\langle2,3\rangle\).

\(\mathbf v(1)=\langle2,3\rangle\)

- Velocity is a vector, while speed is a nonnegative scalar. - Think about how the magnitude of a two-component vector is found. - The sign of a velocity component affects direction but not its squared contribution to speed.

1. Speed is the magnitude of the velocity vector. 2. Thus, \(\|\mathbf v\|=\sqrt{3^2+(-4)^2}=5\).

\(5\)

- Differentiate the position vector once for velocity and twice for acceleration. - Substitute the requested time only after finding the derivative functions. - Compute speed as the nonnegative magnitude of the velocity vector.

1. Differentiate position to get \(\mathbf{v}(t)=\mathbf{r}'(t)=\left\langle 2t, 3\right\rangle\), so \(\mathbf{v}(2)=\left\langle 4, 3\right\rangle\). 2. Differentiate velocity to get \(\mathbf{a}(t)=\mathbf{v}'(t)=\left\langle 2, 0\right\rangle\), so \(\mathbf{a}(2)=\left\langle 2, 0\right\rangle\). 3. The speed is \(\|\mathbf{v}(2)\|=\sqrt{4^{2}+3^{2}}=5\).

Velocity: \(\left\langle 4, 3\right\rangle\); acceleration: \(\left\langle 2, 0\right\rangle\); speed: \(5\)

- Differentiate the coordinate functions once for velocity and again for acceleration. - Use the exact sine and cosine values at \(\frac{\pi}{3}\). - Find speed from the magnitude of the evaluated velocity, not from either component alone.

1. The velocity is \(\mathbf{v}(t)=\left\langle -\sin(t), \cos(t)\right\rangle\), so \(\mathbf{v}\left(\frac{\pi}{3}\right)=\left\langle -\frac{\sqrt{3}}{2}, \frac{1}{2}\right\rangle\). 2. The acceleration is \(\mathbf{a}(t)=\left\langle -\cos(t), -\sin(t)\right\rangle\), so \(\mathbf{a}\left(\frac{\pi}{3}\right)=\left\langle -\frac{1}{2}, -\frac{\sqrt{3}}{2}\right\rangle\). 3. The speed is \(\left\|\mathbf{v}\left(\frac{\pi}{3}\right)\right\|=\sqrt{\frac{3}{4}+\frac{1}{4}}=1\).

Velocity: \(\left\langle -\frac{\sqrt{3}}{2}, \frac{1}{2}\right\rangle\); acceleration: \(\left\langle -\frac{1}{2}, -\frac{\sqrt{3}}{2}\right\rangle\); speed: \(1\)

- Differentiate the position vector once for velocity and twice for acceleration. - Evaluate the derivative vectors at the requested time. - Compute speed from the magnitude of velocity, not acceleration.

1. The velocity is \(\mathbf{v}(t)=\mathbf{r}'(t)=\left\langle e^{t}, 2t\right\rangle\), so \(\mathbf{v}(0)=\left\langle 1, 0\right\rangle\). 2. The acceleration is \(\mathbf{a}(t)=\mathbf{v}'(t)=\left\langle e^{t}, 2\right\rangle\), so \(\mathbf{a}(0)=\left\langle 1, 2\right\rangle\). 3. The speed is \(\|\mathbf{v}(0)\|=\sqrt{1^{2}+0^{2}}=1\).

Velocity: \(\left\langle 1, 0\right\rangle\); acceleration: \(\left\langle 1, 2\right\rangle\); speed: \(1\)

- Simplify or apply the product rule to the first position component before differentiating. - Differentiate once for velocity and a second time for acceleration. - Use the evaluated velocity components to calculate the nonnegative speed.

1. The velocity is \(\mathbf{v}(t)=\left\langle 3t^{2}-3, 2t\right\rangle\), so \(\mathbf{v}(1)=\left\langle 0, 2\right\rangle\). 2. The acceleration is \(\mathbf{a}(t)=\left\langle 6t, 2\right\rangle\), so \(\mathbf{a}(1)=\left\langle 6, 2\right\rangle\). 3. The speed is \(\|\mathbf{v}(1)\|=\sqrt{0^{2}+2^{2}}=2\).

Velocity: \(\left\langle 0, 2\right\rangle\); acceleration: \(\left\langle 6, 2\right\rangle\); speed: \(2\)

- Rewrite reciprocal powers with negative exponents before differentiating if helpful. - Differentiate each component twice, keeping the coordinate order fixed. - Find speed by taking the magnitude of the velocity at \(t=1\).

1. The velocity is \(\mathbf{v}(t)=\left\langle \frac{1}{t}, -\frac{1}{t^{2}}\right\rangle\), so \(\mathbf{v}(1)=\left\langle 1, -1\right\rangle\). 2. The acceleration is \(\mathbf{a}(t)=\left\langle -\frac{1}{t^{2}}, \frac{2}{t^{3}}\right\rangle\), so \(\mathbf{a}(1)=\left\langle -1, 2\right\rangle\). 3. The speed is \(\|\mathbf{v}(1)\|=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}\).

Velocity: \(\left\langle 1, -1\right\rangle\); acceleration: \(\left\langle -1, 2\right\rangle\); speed: \(\sqrt{2}\)

- Differentiate each trigonometric coordinate carefully, including its sign. - Use the exact sine and cosine values at \(t=\pi\). - Calculate speed from the evaluated velocity vector only.

1. The velocity is \(\mathbf{v}(t)=\left\langle 1-\cos(t), \sin(t)\right\rangle\), so \(\mathbf{v}(\pi)=\left\langle 2, 0\right\rangle\). 2. The acceleration is \(\mathbf{a}(t)=\left\langle \sin(t), \cos(t)\right\rangle\), so \(\mathbf{a}(\pi)=\left\langle 0, -1\right\rangle\). 3. The speed is \(\|\mathbf{v}(\pi)\|=\sqrt{2^{2}+0^{2}}=2\).

Velocity: \(\left\langle 2, 0\right\rangle\); acceleration: \(\left\langle 0, -1\right\rangle\); speed: \(2\)

- Displacement is the definite integral of velocity over the time interval. - Integrate the velocity components separately using the same bounds. - Add the displacement vector to the initial position to obtain the final position.

1. The displacement from \(t=0\) to \(t=4\) is \(\int_{0}^{4}\mathbf{v}(t)\,dt=\left\langle 16, 12\right\rangle\). 2. Adding the displacement to the initial position gives \(\mathbf{r}(4)=\left\langle 1, -2\right\rangle+\left\langle 16, 12\right\rangle=\left\langle 17, 10\right\rangle\).

Displacement: \(\left\langle 16, 12\right\rangle\); position at \(t=4\): \(\left\langle 17, 10\right\rangle\)

- Integrate the velocity vector over the interval to find displacement. - Evaluate the sine and cosine antiderivatives at both endpoints. - Add the displacement vector to the given initial position.

1. The displacement is \(\int_{0}^{\pi/2}\mathbf{v}(t)\,dt=\left\langle 1, 1\right\rangle\). 2. Adding displacement to the initial position gives \(\mathbf{r}\left(\frac{\pi}{2}\right)=\left\langle 2, 1\right\rangle+\left\langle 1, 1\right\rangle=\left\langle 3, 2\right\rangle\).

Displacement: \(\left\langle 1, 1\right\rangle\); position at \(t=\frac{\pi}{2}\): \(\left\langle 3, 2\right\rangle\)

- Use the definite integral of velocity from the initial time to the final time. - Integrate the polynomial components separately with bounds \(1\) and \(3\). - Add the resulting displacement to \(\mathbf{r}(1)\) component by component.

1. The displacement is \(\int_{1}^{3}\mathbf{v}(t)\,dt=\left\langle \frac{26}{3}, 8\right\rangle\). 2. Adding displacement to the initial position gives \(\mathbf{r}(3)=\left\langle 0, 5\right\rangle+\left\langle \frac{26}{3}, 8\right\rangle=\left\langle \frac{26}{3}, 13\right\rangle\).

Displacement: \(\left\langle \frac{26}{3}, 8\right\rangle\); position at \(t=3\): \(\left\langle \frac{26}{3}, 13\right\rangle\)

- Integrate both velocity components over \([0,1]\) to obtain displacement. - Preserve the negative sign in the constant vertical velocity. - Add the displacement to the initial vector to find the later position.

1. The displacement is \(\int_{0}^{1}\mathbf{v}(t)\,dt=\left\langle e-1, -1\right\rangle\). 2. Adding displacement to the initial position gives \(\mathbf{r}(1)=\left\langle 1, 4\right\rangle+\left\langle e-1, -1\right\rangle=\left\langle e, 3\right\rangle\).

Displacement: \(\left\langle e-1, -1\right\rangle\); position at \(t=1\): \(\left\langle e, 3\right\rangle\)

- Set the displacement bounds from the given initial time to the requested final time. - Evaluate each polynomial antiderivative carefully at the negative lower bound. - Add the displacement vector to the known position at \(t=-1\).

1. The displacement is \(\int_{-1}^{2}\mathbf{v}(t)\,dt=\left\langle 3, 6\right\rangle\). 2. Adding displacement to the initial position gives \(\mathbf{r}(2)=\left\langle 5, 0\right\rangle+\left\langle 3, 6\right\rangle=\left\langle 8, 6\right\rangle\).

Displacement: \(\left\langle 3, 6\right\rangle\); position at \(t=2\): \(\left\langle 8, 6\right\rangle\)

- Total distance is found by integrating the magnitude of velocity. - Simplify the speed using a fundamental trigonometric identity. - Integrate the resulting constant over the full time interval.

1. The speed is \(\|\mathbf{v}(t)\|=\sqrt{\cos^{2}(t)+\sin^{2}(t)}=1\). 2. Therefore, the total distance is \(\int_{0}^{\pi}1\,dt=\pi\).

\(\pi\)

- A stopped particle must have a zero velocity vector. - Set both velocity components equal to zero independently. - Check that the same parameter value solves both component equations.

1. The particle is stopped when \(\mathbf{v}(t)=\left\langle 0, 0\right\rangle\), so both component equations must hold. 2. Solving \(2t-4=0\) and \(3t-6=0\) gives the common time \(t=2\).

\(t=2\)

- Differentiate the position components to obtain velocity. - Use the signs of the evaluated velocity components to determine horizontal and vertical direction. - Compute speed as the magnitude of the velocity vector.

1. The velocity is \(\mathbf{v}(t)=\left\langle 2t, 3t^{2}-4\right\rangle\). 2. At \(t=1\), \(\mathbf{v}(1)=\left\langle 2, -1\right\rangle\). 3. The positive first component and negative second component mean the particle is moving right and down. 4. Its speed is \(\|\mathbf{v}(1)\|=\sqrt{2^{2}+(-1)^{2}}=\sqrt{5}\).

Moving right and down; speed: \(\sqrt{5}\)

- Compare the endpoint positions before computing distance. - Differentiate the position to find speed. - Constant speed simplifies the distance integral.

1. The initial and final positions are both \((3,0)\), so the displacement is \(\langle0,0\rangle\). 2. The velocity is \(\mathbf{v}(t)=\langle-6\sin(2t),6\cos(2t)\rangle\), so the speed is \(6\). 3. The total distance is \(\int_0^{2\pi}6\,dt=12\pi\).

Displacement: \(\langle0,0\rangle\); total distance: \(12\pi\)

- Directly rightward motion imposes both a zero-component and a sign condition. - Once the time is found, evaluate the velocity magnitude. - Differentiate velocity for the acceleration vector.

1. Directly rightward motion requires the vertical velocity to be \(0\) and the horizontal velocity to be positive. 2. The vertical component is \(t-1\), so \(t=1\). Then the horizontal component is \(2>0\). 3. The speed at that time is \(\|\langle2,0\rangle\|=2\). 4. The acceleration is \(\mathbf{a}(t)=\langle2t,1\rangle\), so \(\mathbf{a}(1)=\langle2,1\rangle\).

At \(t=1\), the particle moves directly right with speed \(2\) and acceleration \(\langle2,1\rangle\).

- Estimate the derivative of velocity using data equally spaced around the target time. - Subtract the earlier velocity vector from the later one. - Divide the entire change vector by the elapsed time.

1. A symmetric estimate is \(\mathbf{a}(2)\approx\frac{\mathbf{v}(2.1)-\mathbf{v}(1.9)}{2.1-1.9}\). 2. The velocity change is \(\langle0.80,0.20\rangle\), and the time change is \(0.20\). 3. Dividing componentwise gives \(\mathbf{a}(2)\approx\langle4,1\rangle\).

\(\mathbf{a}(2)\approx\langle4,1\rangle\)

- Solve the x- and y-coordinate equalities separately; each gives a set of possible times. - A collision must belong to both sets and to the stated time interval. - Evaluate the position vector at every surviving time.

1. Equal x-coordinates require \(t^2=1\), giving the candidate set \(\{-1,1\}\). 2. Equal y-coordinates require \(t=t^3\), so \(t(t^2-1)=0\) and the candidate set is \(\{-1,0,1\}\). 3. The intersection is \(\{-1,1\}\), and both values lie in the stated interval. 4. The collision points are \((1,-1)\) at \(t=-1\) and \((1,1)\) at \(t=1\).

x-coordinate candidates: \(\{-1,1\}\) y-coordinate candidates: \(\{-1,0,1\}\) Collisions: \(t=-1\) at \((1,-1)\), and \(t=1\) at \((1,1)\)

- Compute displacement separately on each time interval. - Direction affects displacement but not the nonnegative distance contribution. - Add vectors for displacement and scalar lengths for distance.

1. From \(t=0\) to \(t=2\), the displacement is \(2\langle1,0\rangle=\langle2,0\rangle\), and the distance is \(2\). 2. From \(t=2\) to \(t=3\), the displacement is \(1\langle-2,0\rangle=\langle-2,0\rangle\), and the distance is \(2\). 3. The total displacement is \(\langle0,0\rangle\), while the total distance is \(4\).

Displacement: \(\langle0,0\rangle\); total distance: \(4\)

- Use the horizontal position coordinate to locate a y-axis crossing. - Apply the time-domain restriction. - Evaluate the signs of both velocity components at the crossing time.

1. The particle is on the y-axis when \(t^2-4=0\). On \(t\ge0\), this gives \(t=2\). 2. The crossing point is \(\mathbf{r}(2)=(0,2)\). 3. The velocity is \(\mathbf{v}(t)=\langle2t,3t^2-3\rangle\), so \(\mathbf{v}(2)=\langle4,9\rangle\). 4. Both components are positive, so the particle is moving right and upward.

At \(t=2\), the particle crosses the y-axis at \((0,2)\), moving right and upward.

- Interpret “horizontal velocity” as a condition on the two components of the velocity vector. - After finding the qualifying time, use the original position function to locate the particle. - Separately verify the maximum claim over the full stated time interval.

1. The velocity is \(\mathbf v(t)=\langle2t-2,-2t+6\rangle\). 2. Horizontal velocity requires the vertical component to be zero while the horizontal component is nonzero. Thus \(-2t+6=0\), so \(t=3\), and \(\mathbf v(3)=\langle4,0\rangle\). 3. The position is \(\mathbf r(3)=\langle3,10\rangle\). 4. The vertical coordinate \(-t^2+6t+1\) is a downward-opening quadratic with its vertex at \(t=3\), so \(10\) is the maximum y-coordinate on \([0,6]\).

At \(t=3\), the velocity is \(\langle4,0\rangle\) and the position is \((3,10)\). The y-coordinate \(10\) is the maximum on \([0,6]\).

- Horizontal motion is determined by the slope of the x-position graph. - Vertical motion is determined by the slope of the y-position graph. - Read the two component slopes at the same time \(t=1\).

1. At \(t=1\), the graph of \(x(t)\) has positive slope, so \(\frac{dx}{dt}>0\) and the particle is moving right. 2. At \(t=1\), the graph of \(y(t)\) has negative slope, so \(\frac{dy}{dt}<0\) and the particle is moving down.

The particle is moving right and down.

- First convert the total displacement from \(A\) to \(B\) into displacement per second. - The position function starts at \(A\) and adds the elapsed-time multiple of the constant velocity. - Use the altitude component to determine the requested time. - For part c), distinguish the velocity vector from its scalar magnitude.

1. The displacement from \(A\) to \(B\) is \(\langle30,30,-30\rangle\). Over \(6\,\text{s}\), the constant velocity is \(\langle5,5,-5\rangle\,\text{m/s}\). 2. Therefore, \(\mathbf{r}(t)=\langle10+5t,20+5t,100-5t\rangle\) for \(0\le t\le6\). 3. Set the altitude equal to \(85\): \(100-5t=85\), so \(t=3\,\text{s}\). Then \(\mathbf{r}(3)=\langle25,35,85\rangle\), giving position \((25, 35, 85)\). 4. Differentiating componentwise gives \(\mathbf{r}'(t)=\langle5,5,-5\rangle\,\text{m/s}\). Its magnitude is \(\sqrt{5^2+5^2+(-5)^2}=5\sqrt{3}\,\text{m/s}\).

a) \(\mathbf{r}(t)=\langle10+5t,20+5t,100-5t\rangle\), \(0\le t\le6\) b) \(t=3\,\text{s}\), at \((25, 35, 85)\) c) Velocity \(\langle5,5,-5\rangle\,\text{m/s}\); speed \(5\sqrt{3}\,\text{m/s}\)

- Total distance is the integral of the velocity vector's magnitude. - Simplify the square root using the relationship between the two velocity components. - Use the interval restriction to resolve the absolute value before integrating.

1. The speed is \(\|\mathbf{v}(t)\|=\sqrt{(3t)^{2}+(4t)^{2}}=5|t|\). 2. Since \(t\ge0\) on the interval, the speed is \(5t\). 3. The total distance is \(\int_{0}^{5}5t\,dt=\frac{125}{2}\).

\(\frac{125}{2}\)

- Form the speed by taking the magnitude of the velocity vector. - Use the nonnegative time interval to simplify the absolute-value factor. - Look for a substitution based on the expression inside the square root.

1. The speed is \(\|\mathbf{v}(t)\|=\sqrt{(2t)^{2}+(t^{2})^{2}}=|t|\sqrt{t^{2}+4}\). 2. Since \(t\ge0\), the speed is \(t\sqrt{t^{2}+4}\). 3. Using \(u=t^{2}+4\), the distance is \(\int_{0}^{2}t\sqrt{t^{2}+4}\,dt=\frac{1}{3}\left[(t^{2}+4)^{3/2}\right]_{0}^{2}=\frac{16\sqrt{2}-8}{3}\).

\(\frac{16\sqrt{2}-8}{3}\)

- Build the speed from both velocity components before integrating. - The resulting definite integral does not need to be simplified to an elementary antiderivative. - Keep full calculator precision until rounding the final distance.

1. The speed is \(\|\mathbf{v}(t)\|=\sqrt{1+\cos^{2}(t)}\). 2. The total distance is \(\int_{0}^{\pi}\sqrt{1+\cos^{2}(t)}\,dt\). 3. Numerical evaluation gives \(3.820197789\ldots\), so the distance is \(\approx3.820\).

\(\approx3.820\)

- Square both velocity components when forming the speed. - Set up the definite integral over the entire time interval before using numerical evaluation. - Round only the final distance, not intermediate values.

1. The speed is \(\|\mathbf{v}(t)\|=\sqrt{e^{2t}+t^{2}}\). 2. The total distance is \(\int_{0}^{1}\sqrt{e^{2t}+t^{2}}\,dt\). 3. Numerical evaluation gives \(1.796429783\ldots\), so the distance is \(\approx1.796\).

\(\approx1.796\)

- A collision time must survive both coordinate equations, not merely one of them. - Keep the two candidate sets separate until each equation has been solved. - Intersect the sets before evaluating a collision point.

1. Equal x-coordinates require \(t=t^2\), so the x-coordinate candidate set is \(\{0,1\}\). 2. Equal y-coordinates require \(t^2=3t-2\), or \((t-1)(t-2)=0\), so the y-coordinate candidate set is \(\{1,2\}\). 3. The intersection of the two candidate sets is \(\{1\}\), so only \(t=1\) satisfies both coordinates. 4. At \(t=1\), both particles are at \((1,1)\).

x-coordinate candidates: \(\{0,1\}\) y-coordinate candidates: \(\{1,2\}\) Common time: \(t=1\) Collision point: \((1,1)\)

- Solve the coordinate equalities independently before deciding that a collision occurs. - A time that works for one coordinate may fail for the other. - The collision times are exactly the intersection of the two solution sets.

1. Equal x-coordinates require \(t+1=t^2+1\), so \(t(t-1)=0\) and the x-coordinate candidates are \(\{0,1\}\). 2. Equal y-coordinates require \(2t=t^2\), so \(t(t-2)=0\) and the y-coordinate candidates are \(\{0,2\}\). 3. Their intersection is \(\{0\}\), so the only collision time is \(t=0\). 4. Both particles are then at \((1,0)\).

x-coordinate candidates: \(\{0,1\}\) y-coordinate candidates: \(\{0,2\}\) Common time: \(t=0\) Collision point: \((1,0)\)

- Match the cosine and sine coordinates to the two coordinates of the fixed point. - Use the stated interval to list only admissible angles. - Keep only an angle that satisfies both trigonometric conditions.

1. A collision requires \(\cos(t)=0\) and \(\sin(t)=1\) at the same time. 2. On \(0\le t<2\pi\), both conditions hold only at \(t=\frac{\pi}{2}\). 3. The common point is \(\left\langle 0, 1\right\rangle\).

The particles collide at \(t=\frac{\pi}{2}\) at the point \(\left\langle 0, 1\right\rangle\).

- Treat the two coordinate equations as independent filters on the same time variable. - Do not discard extra candidates until both equations have been solved. - Evaluate the position only after identifying times common to both sets.

1. Equal x-coordinates require \(t^2=2-t^2\), so \(t=\pm1\) and the x-coordinate candidate set is \(\{-1,1\}\). 2. Equal y-coordinates require \(t=t^2\), so the y-coordinate candidate set is \(\{0,1\}\). 3. Their intersection is \(\{1\}\), so only \(t=1\) is a collision time. 4. The common point is \((1,1)\).

x-coordinate candidates: \(\{-1,1\}\) y-coordinate candidates: \(\{0,1\}\) Common time: \(t=1\) Collision point: \((1,1)\)

- Minimize the squared speed to avoid differentiating a square root. - Express the squared magnitude as a quadratic in \(t\). - Check that the quadratic's minimizing time lies in the stated domain before taking the square root.

1. Since the square-root function is increasing, minimize the squared speed: \(\|\mathbf{v}(t)\|^{2}=t^{2}+(2-t)^{2}=2t^{2}-4t+4\). 2. Completing the square gives \(2(t-1)^{2}+2\), whose minimum on \(t\ge0\) occurs at \(t=1\). 3. The minimum speed is \(\sqrt{2}\).

\(t=1\); minimum speed: \(\sqrt{2}\)

- Work with the square of the speed, which has the same minimizing time as the speed. - Expand or complete the square for the resulting quadratic. - After finding the minimum squared speed, take its nonnegative square root.

1. Minimize the squared speed: \(\|\mathbf{v}(t)\|^{2}=(2t-3)^{2}+(t+1)^{2}=5t^{2}-10t+10\). 2. Completing the square gives \(5(t-1)^{2}+5\), whose minimum on \(t\ge0\) occurs at \(t=1\). 3. The minimum speed is \(\sqrt{5}\).

\(t=1\); minimum speed: \(\sqrt{5}\)

- Identify the type of quantity displacement should be and the type of quantity distance should be. - Ask what the student's value \(14\) is actually doing to horizontal and vertical information. - Compute each requested quantity from its meaning, then compare with the student's claim.

1. The displacement is the vector integral \(\int_{0}^{2}\mathbf{v}(t)\,dt=\left\langle 6, 8\right\rangle\). 2. The speed is the scalar magnitude \(\|\mathbf{v}(t)\|=\sqrt{3^{2}+4^{2}}=5\). 3. The distance is \(\int_{0}^{2}5\,dt=10\). Adding vector components does not produce either displacement or distance.

Displacement: \(\left\langle 6, 8\right\rangle\); distance: \(10\). The value \(14\) incorrectly adds vector components.

- The student's procedure combines net coordinate changes; decide whether that matches Euclidean path length. - Separate the vector question “Where did the particle end relative to where it began?” from the scalar question “How far did it travel?” - Use the geometry of a two-component velocity vector when constructing the scalar travel rate.

1. The displacement is \(\int_{0}^{2}\mathbf{v}(t)\,dt=\left\langle 2, -2\right\rangle\). 2. Adding the absolute values of the net component changes gives a taxicab-style length, not the particle's path length. 3. The speed is \(\|\mathbf{v}(t)\|=\sqrt{t^{2}+(-t)^{2}}=\sqrt{2}\,t\) on \([0,2]\). 4. Therefore, the distance is \(\int_{0}^{2}\sqrt{2}\,t\,dt=2\sqrt{2}\).

Displacement: \(\left\langle 2, -2\right\rangle\); distance: \(2\sqrt{2}\). Adding absolute component changes does not compute Euclidean path length.

- Inspect the student's claimed “speed”: is it a scalar or a vector? - Distinguish the instantaneous magnitude of velocity from the accumulated velocity vector. - Once the correct speed is identified, use it to determine total distance over the interval.

1. The vector \(\int_{0}^{\pi/2}\mathbf{v}(t)\,dt=\left\langle 1, 1\right\rangle\) is the displacement, not the speed. 2. The speed is the scalar \(\|\mathbf{v}(t)\|=\sqrt{\cos^{2}(t)+\sin^{2}(t)}=1\). 3. The distance is \(\int_{0}^{\pi/2}1\,dt=\frac{\pi}{2}\).

Displacement: \(\left\langle 1, 1\right\rangle\); speed: \(1\); distance: \(\frac{\pi}{2}\)

- Minimize the squared distance from the position vector to the origin. - Rewrite the resulting quadratic in vertex form or locate its critical point. - Verify the minimizing time is allowed, then take the square root of the minimum squared distance.

1. Minimize the squared distance \(D^{2}(t)=(t-2)^{2}+(t+1)^{2}=2t^{2}-2t+5\). 2. Completing the square gives \(D^{2}(t)=2\left(t-\frac{1}{2}\right)^{2}+\frac{9}{2}\), so the minimum on \(t\ge0\) occurs at \(t=\frac{1}{2}\). 3. The minimum distance is \(\sqrt{\frac{9}{2}}=\frac{3\sqrt{2}}{2}\).

\(t=\frac{1}{2}\); minimum distance: \(\frac{3\sqrt{2}}{2}\)

- Use the position coordinates to form the square of the distance to the origin. - Minimize the quadratic without introducing a square root. - Apply the domain restriction and then convert the minimum squared distance back to distance.

1. Minimize the squared distance \(D^{2}(t)=t^{2}+(2-t)^{2}=2t^{2}-4t+4\). 2. Completing the square gives \(D^{2}(t)=2(t-1)^{2}+2\), so the minimum on \(t\ge0\) occurs at \(t=1\). 3. The minimum distance is \(\sqrt{2}\).

\(t=1\); minimum distance: \(\sqrt{2}\)

- Approximate horizontal and vertical displacement from their velocity columns separately. - Use the spacing in the time column as the trapezoidal-rule width. - Add the approximate displacement vector to the given starting position.

1. The subinterval width is \(1\). For the horizontal displacement, the trapezoidal rule gives \(\frac{1}{2}[(1+2)+(2+0)+(0-1)]=2\). 2. For the vertical displacement, it gives \(\frac{1}{2}[(0+2)+(2+3)+(3+1)]=\frac{11}{2}\). 3. Thus, the displacement is approximately \(\left\langle 2, \frac{11}{2}\right\rangle\). 4. Adding the initial position gives \(\mathbf{r}(3)\approx\left\langle 4, -2\right\rangle+\left\langle 2, \frac{11}{2}\right\rangle=\left\langle 6, \frac{7}{2}\right\rangle\).

\(\mathbf{r}(3)\approx\left\langle 6, \frac{7}{2}\right\rangle\)

- Use one trapezoidal approximation for each velocity component. - Determine the common subinterval width from consecutive time values. - Interpret the two approximations as displacement and add them to the initial coordinates.

1. The subinterval width is \(0.5\). For the horizontal displacement, the trapezoidal rule gives \(\frac{0.5}{2}[(0+2)+(2+4)+(4+2)]=\frac{7}{2}\). 2. For the vertical displacement, it gives \(\frac{0.5}{2}[(2+4)+(4+2)+(2+0)]=\frac{7}{2}\). 3. Thus, the displacement is approximately \(\left\langle \frac{7}{2}, \frac{7}{2}\right\rangle\). 4. Adding the initial position gives \(\mathbf{r}(1.5)\approx\left\langle 1, 1\right\rangle+\left\langle \frac{7}{2}, \frac{7}{2}\right\rangle=\left\langle \frac{9}{2}, \frac{9}{2}\right\rangle\).

\(\mathbf{r}(1.5)\approx\left\langle \frac{9}{2}, \frac{9}{2}\right\rangle\)

- Apply the trapezoidal rule independently to the horizontal and vertical velocity columns. - Interpret the two numerical integrals as components of displacement. - Add the displacement vector to the position at the initial time.

1. The subinterval width is \(1\). The horizontal displacement is approximately \(\frac{1}{2}[(3+1)+(1+2)+(2+0)]=\frac{9}{2}\). 2. The vertical displacement is approximately \(\frac{1}{2}[(-1+1)+(1+4)+(4+2)]=\frac{11}{2}\). 3. Thus, the displacement is \(\left\langle \frac{9}{2}, \frac{11}{2}\right\rangle\). 4. Adding the initial position gives \(\mathbf{r}(4)\approx\left\langle -2, 5\right\rangle+\left\langle \frac{9}{2}, \frac{11}{2}\right\rangle=\left\langle \frac{5}{2}, \frac{21}{2}\right\rangle\).

\(\mathbf{r}(4)\approx\left\langle \frac{5}{2}, \frac{21}{2}\right\rangle\)

- Horizontal and vertical motion impose different zero-component conditions. - Apply the domain restriction after solving each condition. - Use the sign of the remaining component to determine direction.

1. Horizontal motion requires the vertical component to be \(0\): \(t^2-1=0\). On \(t\ge0\), this gives \(t=1\). 2. At \(t=1\), \(\mathbf{v}(1)=\langle-1,0\rangle\), so the particle moves left. 3. Vertical motion requires the horizontal component to be \(0\): \(t-2=0\), so \(t=2\). 4. At \(t=2\), \(\mathbf{v}(2)=\langle0,3\rangle\), so the particle moves upward.

At \(t=1\), the particle moves horizontally to the left. At \(t=2\), it moves vertically upward.

- A stopped particle has a zero velocity vector, not merely one zero component. - Find the common solution of the two component equations. - Differentiate the velocity vector to obtain acceleration.

1. The particle is stopped when both velocity components are \(0\). 2. The equations \(t^2-4=0\) and \(2t-4=0\) have the common nonnegative solution \(t=2\). 3. The acceleration is \(\mathbf{a}(t)=\mathbf{v}'(t)=\langle2t,2\rangle\). 4. Therefore, \(\mathbf{a}(2)=\langle4,2\rangle\).

The particle is stopped at \(t=2\), and \(\mathbf{a}(2)=\langle4,2\rangle\).

- Relate the rate of change of distance to the position and velocity vectors. - The sign of the relevant dot product determines whether the particle is moving closer or farther away. - Keep the magnitude of the position vector in the denominator.

1. At \(t=1\), the position is \(\langle-3,2\rangle\) and the velocity is \(\langle2,2\rangle\). 2. If \(D=\|\mathbf{r}\|\), then \(D'=\frac{\mathbf{r}\cdot\mathbf{v}}{\|\mathbf{r}\|}\). 3. The dot product is \((-3)(2)+(2)(2)=-2\), and \(\|\mathbf{r}(1)\|=\sqrt{13}\). 4. Thus \(D'(1)=-\frac{2}{\sqrt{13}}\), so the distance is decreasing.

The distance is decreasing at the rate \(\frac{2}{\sqrt{13}}\) units per unit time; equivalently, \(D'(1)=-\frac{2}{\sqrt{13}}\).

- Integrate acceleration to recover velocity first. - Use the initial velocity before integrating a second time. - Apply the initial position after finding the position antiderivative.

1. Integrating acceleration and using \(\mathbf{v}(0)\) gives \(\mathbf{v}(t)=\langle3t^2+1,3-2t\rangle\). 2. Integrating velocity and using \(\mathbf{r}(0)\) gives \(\mathbf{r}(t)=\langle t^3+t-2,3t-t^2+4\rangle\). 3. Evaluating at \(t=2\) gives \(\mathbf{r}(2)=\langle8,6\rangle\).

\(\mathbf{r}(2)=\langle8,6\rangle\)

- Compare squared speeds so that no radical is needed. - Use the speed at the initial time as the threshold. - Intersect the solution of the inequality with the stated time interval.

1. The initial speed is \(\|\mathbf{v}(0)\|=3\). 2. The squared speed is \((t^2-3)^2+4t^2=t^4-2t^2+9\). 3. The condition \(\|\mathbf{v}(t)\|<3\) is equivalent to \(t^4-2t^2<0\), or \(t^2(t^2-2)<0\). 4. On the stated interval, this holds for \(0<t<\sqrt{2}\), whose duration is \(\sqrt{2}\).

The speed is less than its initial speed for \(0<t<\sqrt{2}\). The total duration is \(\sqrt{2}\).

- Read the five values of \(v_x\) from panel a) and the five values of \(v_y\) from panel b). - Apply Simpson’s rule separately to the two component data sets. - Add the resulting displacement vector to the initial position.

1. With \(h=1\), Simpson’s rule gives the horizontal displacement \(\frac{1}{3}[0+0+4(2+2)+2(4)]=8\). 2. The vertical displacement is \(\frac{1}{3}[1+1+4(0+0)+2(-1)]=0\). 3. Adding the displacement \(\langle8,0\rangle\) to the initial position gives \(\mathbf{r}(4)\approx\langle10,-1\rangle\).

\(\mathbf{r}(4)\approx\langle10,-1\rangle\)

- Differentiate velocity to obtain acceleration. - Use a sign test involving velocity and acceleration rather than differentiating a radical. - Evaluate the sign at the specified time.

1. The acceleration is \(\mathbf{a}(t)=\langle1,2t\rangle\). 2. Speed is increasing when \(\mathbf{v}\cdot\mathbf{a}>0\) and decreasing when this dot product is negative. 3. At \(t=\frac{1}{2}\), the dot product is \(\frac{1}{2}+\left(-\frac{3}{4}\right)(1)=-\frac{1}{4}\). 4. Therefore, the speed is decreasing.

The speed is decreasing at \(t=\frac{1}{2}\).

- Differentiate velocity to obtain acceleration. - Use a perpendicular-vector condition. - Apply the positive-time restriction after solving the equation.

1. The acceleration is \(\mathbf{a}(t)=\langle1,2t\rangle\). 2. Perpendicularity requires \(\mathbf{v}(t)\cdot\mathbf{a}(t)=0\). 3. The dot product is \(t+2t(t^2-2)=t(2t^2-3)\). 4. Since \(t>0\), the solution is \(t=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}\).

\(t=\frac{\sqrt{6}}{2}\)

- Use an improper vector integral to find the total displacement. - Add the displacement to the initial position. - Integrate the magnitude of velocity separately for distance.

1. The total displacement is \(\int_0^{\infty}\mathbf{v}(t)\,dt=\langle1,2\rangle\). 2. Therefore, the limiting position is \(\langle1,2\rangle\). 3. The speed is \(\sqrt{e^{-2t}+4e^{-2t}}=\sqrt{5}e^{-t}\). 4. The total distance is \(\int_0^{\infty}\sqrt{5}e^{-t}\,dt=\sqrt{5}\).

Limiting position: \(\langle1,2\rangle\); total distance: \(\sqrt{5}\)

- Horizontal direction depends only on the sign of the horizontal velocity component. - Use its zeros to divide the time domain into intervals. - Test the sign on each interval.

1. Horizontal direction changes can occur when \(v_x(t)=(t-1)(t-3)=0\), so the candidate times are \(t=1\) and \(t=3\). 2. For \(0\le t<1\), both factors are negative, so \(v_x>0\) and the particle moves right. 3. For \(1<t<3\), the factors have opposite signs, so \(v_x<0\) and the particle moves left. 4. For \(t>3\), both factors are positive, so \(v_x>0\) and the particle moves right.

The horizontal direction changes at \(t=1\) and \(t=3\). The particle moves right for \(0\le t<1\) and \(t>3\), and left for \(1<t<3\).

- Work with the square of the speed. - Rewrite it so that one nonnegative trigonometric term controls the minimum. - List every time in the closed interval that produces that minimum.

1. The squared speed is \(S(t)=4\cos^2t+\sin^2t=1+3\cos^2t\). 2. This expression is minimized when \(\cos t=0\). 3. On \([0,2\pi]\), this occurs at \(t=\frac{\pi}{2}\) and \(t=\frac{3\pi}{2}\). 4. The minimum speed is \(\sqrt{1}=1\).

The minimum speed is \(1\), occurring at \(t=\frac{\pi}{2}\) and \(t=\frac{3\pi}{2}\).

- Differentiate velocity to obtain acceleration. - Use a parallel-vector condition in the plane. - Compare the vectors at the valid time to determine orientation.

1. The acceleration is \(\mathbf{a}(t)=\langle1,2t\rangle\). 2. Parallel planar vectors have determinant \(0\): \(t(2t)-(t^2+1)(1)=t^2-1=0\). 3. On \(t\ge0\), the solution is \(t=1\). 4. At \(t=1\), both vectors equal \(\langle1,2\rangle\), so they point in the same direction.

At \(t=1\), the velocity and acceleration vectors are parallel and point in the same direction.

- Differentiate velocity to obtain acceleration. - Relate the derivative of speed to the velocity–acceleration dot product. - Evaluate both the dot product and speed at the stated time.

1. The acceleration is \(\mathbf{a}(t)=\langle3,0\rangle\). 2. The rate of change of speed is \(\frac{\mathbf{v}\cdot\mathbf{a}}{\|\mathbf{v}\|}\). 3. At \(t=2\), \(\mathbf{v}(2)=\langle6,4\rangle\), so \(\mathbf{v}\cdot\mathbf{a}=18\) and \(\|\mathbf{v}\|=2\sqrt{13}\). 4. Therefore, the speed changes at \(\frac{9}{\sqrt{13}}\) units per unit time.

\(\frac{9}{\sqrt{13}}\) units per unit time

- The limiting position depends on the total accumulated displacement. - Evaluate the two improper component integrals separately. - Add the resulting displacement vector to the initial position.

1. The total displacement is \(\int_0^{\infty}\mathbf{v}(t)\,dt\). 2. The horizontal displacement is \(\int_0^{\infty}\frac{2}{(1+t)^2}\,dt=2\). 3. The vertical displacement is \(\int_0^{\infty}-3e^{-t}\,dt=-3\). 4. Adding the displacement to \(\langle2,-1\rangle\) gives the limiting position \(\langle4,-4\rangle\).

\(\langle4,-4\rangle\)

- A circle's radius at the particle points in the direction of the position vector. - A tangent direction is perpendicular to that radius. - Use a vector condition to determine the time before evaluating position.

1. The velocity is \(\mathbf{v}(t)=\langle1,2\rangle\). 2. Tangency to the circle centered at the origin requires the position and velocity vectors to be perpendicular. 3. Solve \(\mathbf{r}(t)\cdot\mathbf{v}(t)=(t-1)+2(2t-4)=5t-9=0\), giving \(t=\frac{9}{5}\). 4. The position is \(\mathbf{r}\left(\frac{9}{5}\right)=\left\langle\frac{4}{5},-\frac{2}{5}\right\rangle\).

\(t=\frac{9}{5}\), at position \(\left\langle\frac{4}{5},-\frac{2}{5}\right\rangle\)

- Recover velocity from acceleration and the initial velocity. - Use a dot-product condition for perpendicular vectors. - Apply the positive-time requirement after solving.

1. Integrating the constant acceleration gives \(\mathbf{v}(t)=\langle1+2t,3-t\rangle\). 2. Perpendicularity requires \(\mathbf{v}(t)\cdot\mathbf{a}(t)=0\). 3. The dot product is \(2(1+2t)-(3-t)=5t-1\). 4. Solving \(5t-1=0\) gives \(t=\frac{1}{5}\).

\(t=\frac{1}{5}\)

- Square the speed equation to remove the radical. - Express the squared speed as the sum of the squared velocity components. - Apply the time-domain restriction after solving the quadratic.

1. The squared speed is \(t^2+(t+2)^2\). 2. Set this equal to \((2\sqrt{2})^2=8\): \(2t^2+4t+4=8\). 3. The resulting equation is \(t^2+2t-2=0\), with solutions \(t=-1\pm\sqrt{3}\). 4. The nonnegative solution is \(t=\sqrt{3}-1\).

\(t=\sqrt{3}-1\)

- Differentiate the position twice. - Compare the acceleration vector directly with the position vector. - Use trigonometric identities to simplify both magnitudes.

1. The velocity is \(\mathbf{v}(t)=\langle-12\sin(3t),12\cos(3t)\rangle\), so the speed is \(12\). 2. The acceleration is \(\mathbf{a}(t)=\langle-36\cos(3t),-36\sin(3t)\rangle\). 3. Since \(\mathbf{a}(t)=-9\mathbf{r}(t)\), acceleration points opposite the position vector and therefore toward the origin. 4. Its magnitude is \(36\).

\(\mathbf{a}(t)=-9\mathbf{r}(t)\), so acceleration always points toward the origin. The speed is \(12\), and the acceleration magnitude is \(36\).

- Read the three horizontal acceleration values from panel a) and the three vertical values from panel b). - Integrating acceleration estimates the change in each velocity component. - Apply Simpson’s rule componentwise, then add the change vector to the initial velocity.

1. With \(h=1\), Simpson’s rule gives the horizontal velocity change \(\frac{1}{3}[0+4+4(2)]=4\). 2. The vertical velocity change is \(\frac{1}{3}[1+1+4(0)]=\frac{2}{3}\). 3. Adding \(\left\langle4,\frac{2}{3}\right\rangle\) to \(\mathbf{v}(0)=\langle1,0\rangle\) gives \(\mathbf{v}(2)\approx\left\langle5,\frac{2}{3}\right\rangle\).

\(\mathbf{v}(2)\approx\left\langle5,\frac{2}{3}\right\rangle\)

- First express speed as the magnitude of the velocity vector. - Use a variable upper limit for accumulated distance. - Solve the resulting equation subject to the time restriction.

1. The speed is \(\sqrt{(2t)^2+(t^2-1)^2}=\sqrt{(t^2+1)^2}=t^2+1\). 2. The distance traveled by time \(T\) is \(\int_0^T(t^2+1)\,dt=\frac{T^3}{3}+T\). 3. Set \(\frac{T^3}{3}+T=\frac{4}{3}\), which gives \(T^3+3T-4=0\). 4. Factor to obtain \((T-1)(T^2+T+4)=0\). The nonnegative real solution is \(T=1\).

\(t=1\)

- Average velocity uses displacement, while average speed uses distance. - Compute the endpoint positions before finding displacement. - The velocity magnitude is constant on this path.

1. The displacement is \(\mathbf{r}(\pi)-\mathbf{r}(0)=\langle-2,0\rangle\). 2. Dividing by the elapsed time \(\pi\) gives the average velocity \(\left\langle-\frac{2}{\pi},0\right\rangle\). 3. The speed is constantly \(1\), so the distance traveled is \(\pi\). 4. Dividing distance by elapsed time gives average speed \(1\).

Average velocity: \(\left\langle-\frac{2}{\pi},0\right\rangle\); average speed: \(1\)

- A collision requires both coordinate equations to hold at the same time. - After finding the common time, evaluate each position to confirm it. - Differentiate both position functions and compare the resulting vectors.

1. Equal x-coordinates require \(t=2-t\), so \(t=1\). 2. At \(t=1\), both y-coordinates equal \(1\), so the particles collide at \((1,1)\). 3. Their velocities are \(\mathbf{v}_1(t)=\langle1,2t\rangle\) and \(\mathbf{v}_2(t)=\langle-1,-2t\rangle\). 4. At the collision, \(\mathbf{v}_1(1)=\langle1,2\rangle\) and \(\mathbf{v}_2(1)=\langle-1,-2\rangle\), so the velocity vectors are opposites.

They collide at \(t=1\) at \((1,1)\). Their velocity vectors are \(\langle1,2\rangle\) and \(\langle-1,-2\rangle\), which are opposites.

- Form the vector from one particle to the other. - Minimize the square of its magnitude instead of the magnitude itself. - Check that the minimizing time lies in the given interval.

1. The separation vector is \(\mathbf{r}_1(t)-\mathbf{r}_2(t)=\langle t,t-2\rangle\). 2. Its squared magnitude is \(D^2(t)=t^2+(t-2)^2=2(t-1)^2+2\). 3. This is minimized at \(t=1\). 4. The minimum separation is \(\sqrt{2}\).

They are closest at \(t=1\), with minimum separation \(\sqrt{2}\).

- Both position coordinates must be zero at the same parameter value. - Check all solutions within the stated interval. - Differentiate the position and evaluate at each visit.

1. The x-coordinate is \(0\) when \(t=\pm1\), and both values also make \(t^3-t=0\). 2. Therefore, the particle visits the origin at \(t=-1\) and \(t=1\). 3. The velocity is \(\mathbf{v}(t)=\langle2t,3t^2-1\rangle\). 4. The two velocity vectors are \(\mathbf{v}(-1)=\langle-2,2\rangle\) and \(\mathbf{v}(1)=\langle2,2\rangle\).

The particle is at the origin at \(t=-1\) and \(t=1\). Its velocity vectors are \(\langle-2,2\rangle\) and \(\langle2,2\rangle\), respectively.

- Form the separation vector before differentiating the distance. - Use the relative velocity, which is the derivative of the separation vector. - The sign of the final scalar rate indicates whether the particles are separating or approaching.

1. The separation vector is \(\mathbf{d}(t)=\mathbf{r}_1(t)-\mathbf{r}_2(t)=\langle-t,t^2+t-4\rangle\). 2. At \(t=1\), \(\mathbf{d}(1)=\langle-1,-2\rangle\), and \(\mathbf{d}'(1)=\langle-1,3\rangle\). 3. If \(D=\|\mathbf{d}\|\), then \(D'=\frac{\mathbf{d}\cdot\mathbf{d}'}{\|\mathbf{d}\|}\). 4. Thus \(D'(1)=\frac{(-1)(-1)+(-2)(3)}{\sqrt{5}}=-\sqrt{5}\).

The distance is decreasing at \(\sqrt{5}\) units per unit time; equivalently, \(D'(1)=-\sqrt{5}\).

- Compare the position and velocity vectors using a dot product whenever the position vector is nonzero. - Factor the resulting expression before making a sign chart. - Check separately what happens when the particle is at the origin.

1. The velocity is \(\mathbf{v}(t)=\langle1,2t\rangle\). 2. When \(\mathbf{r}(t)\ne\langle0,0\rangle\), the sign of \(\mathbf{r}(t)\cdot\mathbf{v}(t)\) determines whether the distance from the origin is decreasing or increasing. 3. The dot product factors as \((t-2)(2t^2+4t+1)\). 4. The quadratic factor is positive for \(t\ge0\), so the particle moves toward the origin for \(0\le t<2\) and away from it for \(t>2\). At \(t=2\), the particle is at the origin and its distance function is not differentiable.

The particle moves toward the origin for \(0\le t<2\) and away from the origin for \(t>2\). At \(t=2\), it passes through the origin, where its distance from the origin is not differentiable.