Find acceleration in this kinematics questions

Question

A car is travelling on a road. The maximum velocity it can attain is 24m/s and the maximum deceleration is 4m/s^2. If the car starts from rest and comes to rest after travelling 1032 m in the shortest time of 56s what is the maximum acceleration that the car can attain?

Answer 1

3 stages of motion: acceleration, constant velocity, deceleration

 

deceleration:

 

v_f=v_i+at

0 m/s = 24 m/s + (-4 m/s²)t

t=6 s

 

Δx =v_i*t + 1/2* a* t²

Δx=(24m/s)(6s)+1/2 *(-4 m/s²)(6s)²

Δx=72 m

 

acceleration and constant velocity

 

time for acceleration t₁ and constant velocity t₂

t1+t2=50

distance for acceleration d1 and constant velocity d2

d1+d2=960

 

d1 = (vi+vf)/2)*t1

d1 = (0+24)/2)*t1

d1 = 12 m/s * t1

 

d2 = 24 m/s * t2

 

960 = 12 m/s * t1 + 24 m/s * t2

960 = 12 m/s *(50-t2) + 24 m/s * (t2)

960 = 12 m/s *t2 + 1200

360 = 12 m/s *t2

t2=30 s

t1=20 s

 

solve for a

vf = vi +at

24 m/s = 0 m/s + a (20 s)

a = 1.2 m/s²