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

A particle moves in the $xy$-plane so that its position is given by the vector $<e^t+t^5,\ln \left(1+5t\right)>$ at time $t\ge 0$.

The acceleration vector of the particle is

A particle travels along a curve in the $xy$-plane such that its velocity vector is $<\frac{3\sqrt{t}}{2},\sin \left(2t\right)+\cos \left(2t\right)>$ where $t\ge 0$. Its position vector at time $t=0$ is $<0,0>$.

The position vector at time $t=\frac{\pi }{2}$ is

If $f$ is a vector-valued function defined by $<\left(t+2\right)e^{-t},e^{-t}+t^2>$, then $f\text{'}\text{'}\left(t\right)$ is

A particle travels along a curve in the $xy$-plane, where its acceleration vector is $<\frac{2}{2t+1},-\frac{2}{{\left(2t+1\right)}^2}>$ at time $t\ge 0$. The velocity vector at time $t=3$ is $<\ln 14,-\frac{1}{7}>$.

The velocity vector at time $t=1$ is

The position of a particle moving in the $xy$-plane is given by the vector $<e^{\sin t},t^2+\cos t>$ where $t$ is time.

The speed of the particle at time $t=2$ is

A particle moves with a position vector of $<\arctan \left(e^t\right),\frac{e^t}{1+e^t}>$ in the $xy$-plane, where time $t\ge 0$. The total distance traveled in the first four seconds is

The acceleration vector of a particle moving in the $xy$-plane is $<2+4\cos \left(2t\right),4-4\sin \left(2t\right)>$, where $t$ is time and the $y$-axis points vertically upwards. At the beginning of its journey, time $t=0$, the particle is at a position vector of $<-2,1>$ moving with a speed of 1 in a direction that is vertically downwards.

The position vector of the particle at time $t=\pi$ is

A particle is moving in the $xy$-plane where the $x$-axis is horizontal and increases towards the right. The position vector of the particle is $<\frac{1}{3}{\left(2t+1\right)}^{\frac{3}{2}}+2\sqrt{1-t},\frac{1}{5}\sqrt{1+t}>$ where $t$ is time and $0<t<1$.

For which of the following time intervals is the particle always moving towards the left?

The function $f$ is a vector-valued function with its second derivative defined by $f\text{'}\text{'}\left(t\right)=<p+qt,p{t}^2>$, where $p$ and $q$ are constants. It is known that $f\text{'}\left(0\right)=<1,-1>$ and $f\text{'}\left(2\right)=<11,7>$ .

The value of $\frac{p}{q}$ is

A particle moves in the $xy$-plane such that, at time $t\ge 0$, its position vector is $<\tan \left(kt\right),e^{kt}-8t^2>$, where $k$ is an integer. If the acceleration vector at time $t=0$ is $<0,0>$ and the $y$-component of the acceleration vector tends towards $-16$ as $t\to \infty$, then