help with physics problem pleaseeeeee?

Question

A car starts from rest and travels for 4.9 s with a uniform acceleration of +1.2 m/s2. The driver then applies the brakes, causing a uniform acceleration of -2.2 m/s2. The breaks are applied for 1.70 s.
How far has the car gone from its start?

Answer 1

First determine the distance the car moves during the period of acceleration.

s = (1/2)at^2
a = 1.2 m/s^2 and t = 4.9 s so:
s = (1/2)1.2(4.9)^2
s = 14.406 meters

Now calculate how fast the car is going at the end of the period of acceleration.

v = at
v = 1.2(4.9) = 5.88 m/s

The driver now applies the brakes and slows down. The distance the car moves during this time is:

s = v0t + (1/2)at^2

Now: a = -2.2 and t = 1.7. v0 is the v from above and = 5.88 m/s
s = 5.88(1.7) - (1/2)2.2(1.7)^2
s = 6 - 3.179
s = 2.821 meters

The distance the car has gone from start is then the sum of the distances during the periods of acceleration and slowing down. so:
Distance = 14.406 + 2.821
Distance = 17.227 meters

Answer 2

HARD