Question

A bus moving at 72 km /hr comes to rest in 20 sec. find the deceleration of the bus

Answer 1

u=72km/hr=20m/sec
v=0km/hr=0m/sec
t=20sec
a=(v-u)/t=(0-20)/20=-20/20=-1m/sec²
Deceleration=1m/sec²
PLEASE MARK AS BRALIEST

Answer 2

The deceleration of the bus = 1 m/s²

Given :

A bus moving at 72 km /hr comes to rest in 20 sec.

To find :

The deceleration of the bus

Solution :

Step 1 of 2 :

Write down given data

Initial velocity

= u

= 72 km/hr

$\displaystyle \sf{ = \frac{72 \times 1000}{3600} \: \: m/s }$

$\displaystyle \sf{ = 20 \: \: m/s }$

Time = t = 20 sec

Final velocity = v = 0

Let acceleration = a m/s²

Step 2 of 2 :

Find deceleration of the bus

By Newton Equation of motion we get

v = u + at

$\displaystyle \sf{ \implies a = \frac{v - u}{t} }$

$\displaystyle \sf{ \implies a = \frac{0 - 20}{20} }$

$\displaystyle \sf{ \implies a = - \frac{20}{20} }$

$\displaystyle \sf{ \implies a = - 1 }$

The deceleration of the bus = 1 m/s²