AP Calculus AB 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals- FRQs - Exam Style Questions

(A) Velocity of \(H\) at \(t=1\)
\(x_H(t)=e^{\,t^{2}-4t}\) so \(v_H(t)=x_H'(t)=e^{\,t^{2}-4t}\times(2t-4)\) by the chain rule (\(\dfrac{d}{dt}e^{u}=e^{u}\times u’\)).
Evaluate at \(t=1\): \(v_H(1)=e^{\,1-4}\times(2\times 1-4)=e^{-3}\times(-2)=\boxed{-2e^{-3}}.\)

(B) Opposite-direction intervals on \(0<t<5\)
For \(H\): \(v_H(t)=(2t-4)\,e^{\,t^{2}-4t}\). Since \(e^{\,t^{2}-4t}>0\) for all \(t\), the sign of \(v_H\) is the sign of \(2t-4\).
Thus \(v_H(t)<0\) on \(0<t<2\) and \(v_H(t)>0\) on \(2<t<5\).
For \(J\): \(v_J(t)=2t\,(t^{2}-1)^{3}\). On \(0<t<1\), \(t>0\) and \(t^{2}-1<0\Rightarrow v_J(t)<0\). On \(1<t<5\), \(t^{2}-1>0\Rightarrow v_J(t)>0\).
Opposite directions occur when one velocity is negative and the other positive: here that happens on \(\boxed{1<t<2}\) (since \(v_H<0\) while \(v_J>0\)).

(C) Is \(J\)’s speed increasing at \(t=2\)?
Speed increases when \(v\) and \(a=v’\) have the same sign (because \(\dfrac{d}{dt}|v|=\text{sgn}(v)\times v’\) when \(v\neq 0\)).
Compute \(v_J(2)=2\times 2\times (2^{2}-1)^{3}=4\times 3^{3}=4\times 27=108>0\). Given \(v_J'(2)>0\).
Since \(v_J(2)>0\) and \(v_J'(2)>0\) have the same sign, \(\boxed{\text{the speed of }J\text{ is increasing at }t=2}.\)

(D) Position of \(J\) at \(t=2\)
With \(x_J(0)=7\), displacement from \(0\) to \(2\) is \(\displaystyle \int_{0}^{2} v_J(t)\,dt=\int_{0}^{2} 2t\,(t^{2}-1)^{3}\,dt\).
Let \(u=t^{2}-1\Rightarrow du=2t\,dt\). When \(t=0\), \(u=-1\); when \(t=2\), \(u=3\).
Then \(\displaystyle \int_{0}^{2} 2t\,(t^{2}-1)^{3}\,dt=\int_{-1}^{3} u^{3}\,du=\left.\frac{u^{4}}{4}\right|_{-1}^{3}=\frac{3^{4}-(-1)^{4}}{4}=\frac{81-1}{4}=20.\)
Therefore \(x_J(2)=x_J(0)+\text{displacement}=7+20=\boxed{27}.\)