AP Calculus AB : 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration- Exam Style questions with Answer- FRQ

Ans:

2(a) There is one time, t = 

, in the interval 0 < 7 < 2 when the particle is at rest (not moving). Find 

.For 0 <t < 

, is the particle moving to the right or to the left? Give a reason for your answer.

v(t) =0 

t = 1.425610

Therefore, the particle is at rest (not moving) at 

, = 1.426 (or 1.425).

For 

. Therefore, the particle is moving to the right on that interval.

2(b) Find the acceleration of the particle at time t = 1.5. Show the setup for your calculations. Is the speed of the particle increasing or decreasing at time t = 1.5 ? Explain your reasoning.

a(1.5) =v'(1.5) = -1

The acceleration of the particle at time t = 1.5 is —1 (or —0.999).

v(1.5) = —0.076856 < 0

Because a(1.5) and v(1.5) have the same sign, the speed is increasing at time t = 1.5.

2(c) The position of the particle at time t is x(t), and its position at time t = 1 is x(1) = —3. Find the position of the particle at time t = 4. Show the setup for your calculations.

$x\left(4\right)=x\left(1\right)+{\int }_1^{4}v\left(t\right)dt$

= -3+0.197117 = -2.802883

The position of the particle at time t = 4 is —2.803 (or —2.802 ).

2(d) Find the total distance traveled by the particle over the interval 

. Show the setup for your calculations.

= 0.9581

The total distance traveled by the particle over the interval 

is 0.958.