Question

A Jeep starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110m. Determine the acceleration of the Jeep.

Answer 1

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

Given that,

Initial speed of a jeep, u = 0

Time, t = 5.21 s

Distance, d = 110 m

To find,

The acceleration of the jeep.

Solution,

To find the acceleration, in this case, use second equation of kinematics as :

$d=ut+\dfrac{1}{2}at^2$

Put all values, we get :

$d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 110}{(5.21)^2}\\\\a=8.1\ m/s^2$

So, the acceleration of the jeep is $8.1\ m/s^2$