Question

8. A car starts from rest and moves with constant
acceleration of 4 m/s2 for 30 seconds. Then the
brakes are applied and the car comes to rest in
another 60 seconds. Find [5]
(i) Maximum velocity attained by car
(ii) Magnitude of retardation of car
(iii) Total distance covered by car
(iv) Average speed of car through out the motion

Answer 1

Answer:

(i) 120 m/s

(ii) -2 m/s²

(iii) 5400 m

(iv) 60 m/s

Explanation:

For (i), Max velocity will be Attained just before Retardation,

Applying Equations of Motion, v = u + at

v = 0 + 4 x 30 = 120 m/s

(ii) Final Velocity is 0, so v = 0 m/s and At the Start of retardation u = 120 m/s

so By v = u + at

0 = 120 + a x 60

a = -2 m/s²

(iii) We distribute the trip distance into 2 parts, s1 = before Retardation and s2  = after Retadation  

By, s = ut + 1/2 at²

s1 = 0t + 1/2 x 4 x (30)²

s1 = 1800 m

For s2 = 120 x 60 + 1/2 x (-2) x 60²

s2 = 3600 m

Total Dist = s1 + s2 = 5400 m

For (iv) We know Avg. Speed = Total Dist / Total Time

Avg. Speed = 5400/30+60 = 60 m/s