Question

A bus starting from rest moves with a uniform acceleration of 0.2 ms- ^2 for 120 s.Find the speed acquired by the bus.​

Answer 1

GIVEN:-

• $\rm{Initial\:Velocity = 0m/s^{-1}}$

• $\rm{Acceleration = 0.2m/s^{-2}}$

• $\rm{Time\:Taken = 120s.}$

TO FIND:-

• The speed acquired by bus i.e Final Velocity.

FORMULAE USED:-

• ${\huge{\boxed{\rm{v = u + at}}}}$

Where,

v = Final Velocity

u = Initial Velocity

a = Acceleration

t = Time.

Now,

$\implies\rm{V = u + at}$

$\implies\rm{V = 0 + 0.2\times{120}}$

$\implies\rm{V = 24m/s^{-1}}$

Hence, The Speed acquired by bus is 24m/s¹.

Kinematics Equation.

• ${\huge{\boxed{\rm{v^2 - u^2 = 2as}}}}$

• ${\large{\boxed{\rm{s = ut + \dfrac{1}{2}\times{a}\times{(t)^2}}}}}$>

Answer 2

Explanation:

Given; acceleration is 0.2 m/s², time is 120 sec and initial velocity is 0 m/s (as the bus starts from rest).

To find: final velocity (speed) of the bus.

First equation of motion, v= u + at

Second equation of motion, s = ut + 1/2 at²

Third equation of motion, v² - u² = 2as

As we have value of u, a and t. So,

Using the first equation of motion,

v = u + at

Substitute the values,

→ v = 0 + 0.2(120)

→ v = 0 + 24.0

→ v = 24

Hence, the final velocity of the bus is 24 m/s.