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

`d=109.2m`

Step-by-step 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+(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+(a_1t_1^2)/(2)=(0m/s)(6s)+((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=(v_3^2-v_(i3)^2)/(2a_3)=((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`