Question

A bus travelling at a speed of 18km/h. it stopped in 2.5second by applying break.find (1.)acceralation due to brake(2.) distance covered by bus.?​

Answer 1

Answer:

Given u=18km/hr

1km/hr=5/18m/s

18km/hr=18×5/18m/s

u=5m/s

t=2.5sec

v=0

From 1st equations of motion

v=u+at

0=5+a (2.5)

-5=a (2.5)

a=-2m/s2

From second equations of motion

$s = ut + \frac{1}{2} a {t}^{2}$

s=5×2.5+1/2×-2×2.5×2.5

s=6.25

Answer 2

Answer:-

a = -2.5m/s²

s = 6.25m

Given :-

u = 18 km/hr

= 18 × 5/18

= 5 m/s

t = 2.5 s

v = 0m/s

To find:-

1)The acceleration due to brake.

2) Distance covered by bus.

Solution:-

Since,the bus applied the break and come to rest.

Deaceleration produced will be given by :-

$\huge \boxed {a = \dfrac{v-u}{t}}$

Put the given value ,

→$a = \dfrac{0-5}{2.5}$

→$a = \dfrac{-5}{2.5}$

→$a = -2 m/s^2$

Now , we have

a = -2m/s²

Then,

The distance covered by the car is given by :-

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

→$2 \times -2 \times s = (0)^2-(5)^2$

→$-4s = -25$

→$s = \dfrac{-25}{-4}$

→$s = 6.25 m$

hence,

The distance travelled by bus is 6.25m.