Math  /  Calculus

QuestionJ Test Booklet
FRQ 4.1 and 4.2 hegafice (gany buck herve) Caren rides her bicycle along a straight road from home to school, starting at home at time t=0t=0 minutes and arriving at school at time t=12t=12 minutes. During the time interval 0t120 \leq t \leq 12 minutes, her velocity v(t)v(t), in miles per minute, is modeled by the piecewise-linear function whose graph is shown above.
2. Shortly after leaving home, Caren realizes she left her calculus homework at home, and she returns to get it. At what time does she turn around to go back home? Give a reason for your answer. Canen turns aroun I to go hame at t=2t=2 minutes beracde at theat tiphe we can see that her nelocity deereazes arich cun mean she went back hone. A particle moves along a straight line. For 0t50 \leq t \leq 5, the velocity of the particle is given by v(t)=2+(t2+3t)6/5t3v(t)=-2+\left(t^{2}+3 t\right)^{6 / 5}-t^{3}, and the  position of the particle is given by s(t). It is known that s(0)=10.2+(t2+3t)6/513=2v(t)=22+(t2+3t)6/5t3=2(t2+3t)6/5t3=0(t2+3t)6/s=t33t)6/5t3=4\begin{array}{l} \text { position of the particle is given by } s(t) \text {. It is known that } s(0)=10 . \quad-2+\left(t^{2}+3 t\right)^{6 / 5}-1^{3}=2 \\ v(t)=-2 \\ -2+\left(t^{2}+3 t\right)^{6 / 5}-t^{3}=-2 \\ \left.\left(t^{2}+3 t\right)^{6 / 5}-t^{3}=0 \quad\left(t^{2}+3 t\right)^{6 / s}=t^{3} \quad-3 t\right)^{6 / 5}-t^{3}=4 \end{array}
3. 国 Find all values of tt in the interval 2t42 \leq t \leq 4 for which the speed of the particle is 2 . V(t)=2|V(t)|=2 in the interval 2ty2 \leq t \leq y are t2.80t \approx 2.80 and t3.292t \approx 3.292. a(t)=v(t)=sin(t22)(t+1)cos(t22)ta(t)=v(t)=-\sin \left(\frac{t^{2}}{2}\right)-(t+1) \cos \left(\frac{t^{2}}{2}\right) \cdot t

A particle moves along the xx-axis so that its velocity at time tt is given by a(t)=sin(t22)t(t+1)cos(t22)a(t)=-\sin \left(\frac{t^{2}}{2}\right)-t(t+1) \cos \left(\frac{t^{2}}{2}\right) v(t)=(t+1)sin(t22),a(2)=sin(222)2(21)cos(z22)v(2)=(2+1)sin(222)=3sin(2)=sin(2)6cos(2)\begin{array}{l} v(t)=-(t+1) \sin \left(\frac{t^{2}}{2}\right), a(2)=-\sin \left(\frac{2^{2}}{2}\right)-2(2-1) \cos \left(\frac{z^{2}}{2}\right) \\ v(2)=-(2+1) \sin \left(\frac{2^{2}}{2}\right)=-3 \sin (2)=-\sin (2)-6 \cos (2) \end{array}
At time t=0t=0, the particle is at position x=1x=1.
4. 园 Find the acceleration of the particle at time t=2t=2. Is the speed of the particle increasing at t=2t=2 ? Why or why not? The ucueteration at y=2y=2 is Page 2 of 2 AP Calculus AB

Studdy Solution
Determine if the speed is increasing at t=2 t = 2 by checking the signs of v(2) v(2) and a(2) a(2) . If both have the same sign, the speed is increasing.
The solution to the problems are:
1. Caren turns around at t=2 t = 2 minutes.
2. The acceleration of the particle at t=2 t = 2 is calculated from the derived function.
3. The speed of the particle is increasing at t=2 t = 2 if both velocity and acceleration have the same sign.

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