Question

1::: A jeep starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 meter.Determine the acceleration of the Jeep?

2::::A truck starting from rest moves with a uniform acceleration of 0.2 m/s^2 for 2 minutes ,.Find (a) the speed acquired
(B) the distance travelled

Answer 1

1.
S = ut + 0.5at²
a = 2S / t²
= 2 × 110 / (5.21)²
= 8.10 m/s^2
___________________

2.
(a)
v = u + at
= 0 + (0.2 × 2 × 60)
= 24 m/s
Speed acquired is 24 m/s

(B)
S = ut + 0.5at^2
= 0 + (0.5 × 0.2 × (2 × 60)^2)
= 1440 m
= 1.44 km
Distance travelled is 1.44 km

Answer 2

Answer:

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