Question

A bus moving with a velocity of 15 m/s please stop the bus by applying brakes to provide a uniform declaration of
of 5m /S2 calculate the distance travelled by bus before ​

Answer 1

Answer: here final v=0

initial v=15

a= -5

hence we will apply

v^2 - u^2=2as where v=0 u=15 a=-5 and s=?

0^2 - (15)^2=2*(-5)s

on solving

we have

s=22.5 meters

Answer 2

Given :-

★ Initial velocity, u = 15 m/s

★ Acceleration, a = 5 m/s²

★ Final velocity , v = 0 m/s

( As break applied)

To Find :-

★ Distance, s

Solution :-

To find s ( Distance) , we need to use 3rd equation of motion.

We know,

$\sf v^2-u^2=2as$

Where,

v = Final velocity

u = Initial velocity

a = Acceleration

s = Distance

Substituting the values, we get __

⟹ ( 0)² - (15) ² = 2 × 15 × s

⟹ s = 22.5 m

Hence,

the distance covered by bus is = 22.5 m