Question

A car starts from rest at a stop sign. It accelerates at 2.0 m/s^2 for 6.0 seconds, coasts for 2.1 s , and then slows down at a rate of 1.5m/s^2 for the next stop sign. How far apart are the stop signs?

Answer 1

Answer:

$d=109.2m$

Explanation:

We divide the problem in 3 parts: part 1 when the car accelerates, part 2 when its velocity is constant and part 3 when it slows down and we use the main equations for accelerated motion $v=v_i+at$ and $d=v_it+\frac{at^2}{2}$, and for the last part $v^2=v_i^2+2ad$

We start calculating the displacement and final velocity of part 1 considering the initial velocity is zero:

$d_1=v_{i1}t_1+\frac{a_1t_1^2}{2}=(0m/s)(6s)+\frac{(2m/s^2)(6s)^2}{2}=36m$

$v_1=v_{01}+a_1t_1=(0m/s)+(2m/s^2)(6s)=12m/s$

Now we calculate the displacement of part 2 considering the velocity remains constant since acceleration is zero:

$d_2=v_{i2}t_2=(12m/s)(2.1s)=25.2m$

Now we calculate the displacement of part 3 considering the initial velocity is the final velocity of the previous part and that it slows down until coming to rest:

$d_3=\frac{v_3^2-v_{i3}^2}{2a_3}=\frac{(0m/s)^2-(12m/s)^2}{2(-1.5m/s^2)}=48m$

So our total displacement is $d=d_1+d_2+d_3=36m+25.2m+48m=109.2m$