Question

A car starts from rest, then accelerates at 1.20 m/s^2 for 7.00 s. It hits the brakes, slowing to stop at a rate of -4.25 m/s^2. What is the total time for the problem? (Unit = s) PLEASE HELP!!

Answer 1

Answer:

8.98 secs

Explanation:

First part of the journey;

If the car starts from rest, then accelerates at 1.20 m/s^2 for 7.00 s, the total time taken during this period is 7.0s.

But then  lets get the velocity v of the car during its first part of the journey.

Using the formula v1 = u1+ at1

v = 0+1.2(7)

v = 8.4m/s

Second part of the journey is when the driver hits the brake and slowing to stop at a rate of -4.25 m/s^2. To calculate the time it takes by the body to decelerates, we will use the equation of motion v2 = u2 + at2

v = final velocity of the car during the second part of the journey

u = initial velocity of the car during the second part of the journey

a = deceleration

t = time taken

Given v2 = 0m/s, u2 = 8.4m/s, a = -4.25m/s²

0 = 8.4-4.25t2

4.25t2 = 8.4

t2 = 8.4/4.25

t2 = 1.98secs

Total time taken for the journey = t1+t2

= 7.0s + 1.98s

= 8.98secs