Question

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

Answer 1

Answer:

The acceleration of the jeep is 8.1 $\sf \: m/s^{2}$

GIVEN :

Initial speed of jeep, u = 0

Time, t = 5.21 sec.

Distance, d = 110 m.

TO FIND :

The acceleration of the jeep = ?

FORMULA :

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

SOLUTION :

Using equation of motion,

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

→ $\sf d \: = \: \dfrac {1}{2} at^{2}$

→ $\sf a \: = \: \dfrac {2d}{t^{2}}$

→ $\sf a \: = \: \dfrac {2 \times 110}{(5-1)^{2}}$

$\Rightarrow \sf \red{\boxed{\bold{a \: = \: 8.1 \: m/s^{2}}}}$

$\therefore$ The acceleration of the jeep is $\sf 8.1 \: m/s^{2}$.

Second equation of motion :-

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

• Second equation of motion is used to find the distance, time and acceleration.