Vector-Valued Functions (College Board AP® Calculus BC): Exam Questions

A particle moves in the $xy$-plane with an acceleration vector of $<6e^{2t}+\sin t,6e^{-2t}-\sin t>$ at time $t\ge 0$. At time $t=0$ the velocity vector is $<-8,4>$.

Find the velocity vector in terms of time, $t$.

The position vector of a particle moving in the $xy$-plane at time $t\ge 0$ is $<\frac{1}{3t+1},4t^{\frac{3}{2}}>$.

Find the speed of the particle at time $t=0.4$.

The position vector of a particle moving along a curve in the $xy$-plane at time $t>1$ is $<\sin \left(\frac{1}{2}t\right),t\ln t>$.

Find the acceleration vector at time $t=\pi$.

For time $t\ge 0$, a particle is moving in the $xy$-plane with a position vector of $<t^3,\cos \left(1+t^2\right)>$.

Find the total distance traveled by the particle over the time interval $0\le t\le 2.5$.

At time $t\ge 0$, a particle moving along a curve in the $xy$-plane has a position vector of $<5t^4-t,10e^{t^3}>$.

For $0<t<1$, there is a point on the curve at which the line tangent to the curve has a slope of $25$. Find the time at which the object is at this point.

A boat is moving on the surface of the sea. At time $t\ge 0$, the position of the boat is the point with coordinates $\left(x\right(t),y(t\left)\right)$, where $x\text{'}\left(t\right)=662\text{sin}\left(5t\right)$ and $y\text{'}\left(t\right)=880\text{cos}\left(6t\right)$. Time $t$ is measured in hours, and $x\left(t\right)$ and $y\left(t\right)$ are measured in meters. Find the total distance traveled by the boat over the time interval $0\le t\le 1$.

At time $t\ge 0$, a particle moves in the $xy$-plane with position $\left(x\left(t\right),y\left(t\right)\right)$ and an acceleration vector of $<20t^3+8,e^t-2>$. It is known that, at time $t=0$, the particle is at a position of $\left(1,3\right)$ with a velocity vector of $<-1,4>$.

Find the position vector of the particle in terms of time, $t$.

For time $t\ge 0$, a particle is moving in the $xy$-plane along a curve with a position vector of $<\left(2t-5\right)e^{2t},6t^3+6>$.

Find the coordinates of the point on the curve that is the farthest to the left.

Explain why there is no point on the curve that is farthest to the right.

A particle moves in the $xy$-plane with an acceleration vector of $<3t^2\cos \left(t^3\right),3t^2\sin \left(t^3\right)>$ at time $t\ge 0$. At time $t=0$, the velocity vector is $<1,0>$.

The total distance traveled by the particle in the interval $0\le t\le 10$ is half the total distance traveled by the particle in the interval $10<t\le m$. Write, but do not solve, an equation involving one or more integrals whose solution gives the value of $m$.

At time $t\ge 0$, a particle is moving along a curve in the $xy$-plane with a position vector of $<at+b+\cos \left(3t\right),pt+q+\sin \left(3t\right)>$ where $a$, $b$, $c$ and $d$ are constants.

Explain why the magnitude of the acceleration never changes throughout the motion of the particle, for $t\ge 0$.

The velocity vector of a particle moving along a curve in the $xy$-plane at time $t$ where $0\le t<\frac{\pi }{2}$ is $<\frac{1}{1+t^2},\frac{e^t}{1+e^t}>$. The position vector of the particle at time $t=0$ is $<-1,\ln 2>$.

Show that an equation of the curve followed by the particle is $e^y=1+e^{\tan \left(x+1\right)}$.

For time $t\ge 0$, a particle moves in the $xy$-plane with position $\left(x\right(t),y(t\left)\right)$ and velocity vector $\left((t-1)e^{t^2},\text{sin}\left(t^{1.25}\right)\right)$. At time $t=0$, the position of the particle is $(-2,5)$.

Find the speed of the particle at time $t=1.2$. Find the acceleration vector of the particle at time $t=1.2$.

Find the total distance traveled by the particle over the time interval $0\le t\le 1.2$.

Find the coordinates of the point at which the particle is farthest to the left for $t\ge 0$. Explain why there is no point at which the particle is farthest to the right for $t\ge 0$.