Question

A truck is moving with the speed of 72 km/h. Suddenly, driver
sees an obstacle and he applies the brake to stop the truck in
5 sec. Find the retardation produced by brakes and the
distance travelled by the truck before it stops.​

Answer 1

_: ANSWER :_

Given : -

A truck is moving with the speed of 72 km/hr . Suddenly , driver sees an obstacle and he applies the brake to stop the truck in 5 seconds .

Required to find : -

• Retardation of the truck ?

• Distance travelled by the truck before it stops ?

Equations used : -

❶ v = u + at

❷ v² - u² = 2as

Here,

• v = Final velocity

• u = Initial velocity

• a = acceleration

• t = time taken

• s = displacement

Solution : -

A truck is moving with the speed of 72 km/hr . Suddenly , driver sees an obstacle and he applies the brake to stop the truck in 5 seconds .

we need to find the ;

• Retardation of the truck ?

• Distance travelled by the truck before it stops ?

So,

From the given information we can conclude that ;

• Initial velocity of the truck ( u ) = 72 km/hr

• Time ( t ) = 5 seconds

• Final velocity of the truck ( v ) = 0 km/hr

Here,

we need to convert the initial & final velocities from km/hr into m/s because the SI unit of acceleration is m/s² .

This implies ;

✮ 1 km/hr = 5/18 m/s

↪ 72 km/hr = ?

↪ 72 x 5/18

↪ 4 x 5

↪ 20 m/s

72 km/hr = 20 m/s

Similarly,

0 km/hr = 0 m/s

Hence,

• Initial velocity of the truck ( u ) = 20 m/s

• Final velocity of the truck ( v ) = 0 m/s

Now,

Let's find the acceleration of the truck .

Using the equation of motion ;

i.e. v = u + at

↪ 0 = 20 + a x 5

↪ 0 = 20 + 5a

↪ 0 - 20 = 5a

↪ - 20 = 5a

↪ 5a = - 20

↪ a = - 20/5

↪ a = - 4 m/s²

Since,

Negative acceleration is said to be as Retardation .

So,

• Retardation of the truck = - 4 m/s²

Now,

Let's find the displacement of the truck .

Using the equations of motion ;

i.e. v² - u² = 2as

↪ ( 0 )² - ( 20 )² = 2 x - 4 x s

↪ 0 - 400 = 2 x - 4 x s

↪ - 400 = 2 x - 4 x s

↪ - 400 = - 8 x s

↪ - 400 = - 8s

Taking - ( minus ) common on both sides

↪ - ( 400 ) = - ( 8s )

- ( minus ) get's cancelled on both sides

↪ 400 = 8s

↪ 8s = 400

↪ s = 400/8

↪ s = 50 meters

Hence,

• Displacement of the truck ( s ) = 50 meters

Since,

Displacement can be taken or consider in terms of distance if required .

So,

• Distance travelled by the truck ( s ) = 50 meters

Answer 2

acceleration = - 4 m/s²

Distance traveled = 50 meters

Using the Equations of motion ;

i.e.

v = u + at

s = ut + ½ at²