Question

A jeep starts from rest and accelerates Uniformly over a time of 5.21 sec for a distance of 110 m determine accleration

Answer 1

Answer:

Explanation:

Use equation of motion

S = ut + 0.5at^2

110 = 0 + [0.5a × (5.21)^2]

a = 110 / (5.21^2 × 0.5)

a = 8.10 m/s^2

Acceleration of jeep is 8.10 m/s^2

Answer 2

Answer:21.1m/s

Explanation:v-u/t=110-0/5.21 = 110/5.21 = 21.1m/s