Question

A bus is moving with a velocity of 72 Kmh-1. On the application of brakes, it
stops after covering a distance of 500m. Calculate the deceleration produced by
the brakes.​

Answer 1

Explanation:

u=72kmph

72kmph=20m/s

s=500m

v=0

substitute the values in the equation: v^2-u^2=2as

0-400=2(a)500

so, retardation=0.4m/sec^2

hope it helps you

Answer 2

Given :-

A bus is moving with a velocity of 72 km/hr .

On applying brakes the bus stops after covering a distance of 500 meters .

Required to find :-

• Deceleration of the bus produced by the brakes ?

Concept used :-

• Equations of motion

Equation used :-

$\large{\dagger{\boxed{\rm{ {v}^{2} - {u}^{2} = 2as }}}}$

Units Conversion :-

Here we need to convert the velocity from meters to kilometres

Reason :-

Since the displacement is given in km/hr we need to convert the velocity into metres per second .

1 km/hr = 5/18 m/s

72 km/hr = ? m/s

=> 72 x 5/18

=> 20 m/s

This implies,

Velocity of the bus = 20 m/s

Similarly,

0 km/hr = 0 m/s

Solution :-

Given that :-

A bus is moving with a velocity of 72 km/hr ( 20 m/s )

on applying brakes the bus stops after covering a distance of 500 meters

So,

From the above we can conclude that ;

• Initial velocity of the bus = 20 m/s

• Final velocity of the bus = 0 m/s

• Displacement = 500 meters

We need to find the deceleration of the bus

So,

Using the 3rd equation of motion

$\large{\dagger{\boxed{\rm{ {v}^{2} - {u}^{2} = 2as }}}}$

Here,

• v = Final velocity

• u = Initial velocity

• a = acceleration

• s = displacement

Since negative acceleration is known as retardation or deceleration .

So,

By substituting the values in the equation

We get,

$\longrightarrow{\tt{ {(0\;m/s)}^{2} - {(20\;m/s)}^{2} = 2 \times a \times 500 }}$

$\longrightarrow{\tt{ 0 \; m^2/s^2 - 400 \; m^2/s^2 = 1000a }}$

$\longrightarrow{\tt{ - 400 \; m^2/s^2 = 1000a }}$

$\longrightarrow{\tt{ 1000a = - 400 \; m^2/s^2}}$

$\longrightarrow{\tt{ a = \dfrac{ - 400 \; m^2/s^2 }{ 1000 \; m }}}$

$\longrightarrow{\tt{ a = \dfrac{ -4 \; m/s^2 }{10 }}}$

$\longrightarrow{\tt{ a = - 0.4 \; m/{s}^{2}}}$

Therefore ,

Deceleration of the bus = - 0.4 m/s²