A car starts from rest and travels for 5.0 s with a uniform acceleration of + 1.5 m/s2. The driver then applies the brakes, causing a uniform acceleration of -2.0 m/s2. If the brakes are applied for 3.0 s, calculate (a) how fast is the car going at the end of the braking period, and (b) how far has it gone?

Answer 1

v = Vi + a t
x = X1 + Vi t +(1/2) a t^2
phase1
Xi = 0
Vi = 0
when t = 5 with a = 1.5
v = 0 + 5 t = 5*1.5 = 7.5 m/s
x = 0 + 0 + (1/2) (1.5) (25) = 18.75 meters
Phase 2
Xi = 18.75
Vi = 7.5
a = -2
t = 3
v = 7.5 - 2* 3 = 1.5 m/s
x = 18.75 + 7.5 *3 -(1/2)(2)(9)
= 18.75 + 22.5 - 9
= 13.5

Answer 2

Phase 1
Use "vf=a(t)+vi" to find final velocity of phase 1, which is also the initial velocity of phase 2. Then use "x=1/2(vi+vf)t" to find displacement of first phase, which will be added later to displacement of second phase.
a = 1.5 m/s^2
t = 5.0 s
vi = 0 m/s
vf = 1.5 (5.0) + 0 = 7.5 m/s
x = 1/2 (0 + 7.5) (5.0) = 18.75 m

Phase 2
Use the final velocity found in phase 1 as the initial velocity of this phase. "vf=a(t)+vi" will help to find the final velocity after the braking period. Then use the final velocity to calculate the displacement of the second phase. Add the displacements of the two phases together.
a = -2.0 m/s^2
t = 3.0 s
vi = 7.5 m/s
vf = -2 (3.0) + 7.5 = 1.5 m/s
x = 1/2 (7.5 + 1.5) (3.0) = 13.5 m
18.75 m + 13.5 m = 32.25

The car is going 1.5 m/s at the end of the braking period. The car has travelled 32.25 m from its start.