Question

A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.7 seconds, coasts for 2.8 s , and then slows down at a rate of 1.5m/s2 for the next stop sign. PART A : How far apart are the stop signs? (Please explain how you got the answer too, thanks:))

Answer 1

Answer:

18.43m

Explanation:

At the initial stage, the car accelerates at 2.0 m/s² for 6.7 seconds, to get the velocity of the car at this point, we will use the equation of motion;

v = u + at

v = 0+(2.0)(6.7)

v = 13.4m/s

The velocity of the car during this time is 13.4m/s.

Also, if the car slows down at a rate of 1.5m/s², we can also calculate the time it took for the car to decelerate (check the attachment for diagram).

v = u+at

v = 0

u = 13.4m/s

a = -1.5m/s² (deceleration is negative acceleration)

0 = 13.4+-1,5t

-13.4 = -1.5t

t = 13.4/1.5

t = 8.93s

Hence it took the car 8.93s to slow down for the next stop sign.

To calculate how far apart the stop signs are, we need to calculate the total distance AD according to the diagram

Distance covered = AD = 18.43m

HENCE THE STOP SIGNS ARE 18.43m apart