Question

A bus is travelling at a speedof 54 km/h . Brakes are appliedso as to produce a uniformacceleration of - 0.8 m/s*-2. Findhow far the train will go before itis brought to rest.

Answer 1

Answer:

Before coming to rest, bus covers a distance of 141 m.

Explanation:

= > 54 km / hr

= > 54 { 1000 m / ( 60 x 60 sec ) }

= > 54 ( 5 / 18 ) m/s

= > 15 m/s

Here,

Initial velocity( u ) of bus is 15 m/s.

Uniform acceleration( retardation ) acting opposite to the bus is - 0.8 m/s^2 .

Final velocity of bus is 0.

Using the third equation to motion ( velocity-displacement equation ) :

• v^2 = u^2 + 2aS, where symbols have their usual meaning

Here,

= > 0 = ( 15 m/s )^2 - 2( 0.8 m/s^2 ) S

= > 0 = 225 m - 1.6S

= > 225 / 1.6 m = S

≈ 141 m = S

Before coming to rest, bus covers a distance of 141 m.

Answer 2

Answer:

140.625 m

Explanation:

Given :

Initial velocity = 54 km / hr or 15 m / sec

When break is applied v = 0 m / sec

Acceleration = - 0.8 m / sec²

We have to find distance i.e. s

We have :

v² = u² + 2 a s

0² = u² + 2 a s

- u² = 2 a s

- 15² = 2 × ( -0.8 ) × s

s = -225 / -1.6

s = 140.625 m

Hence , the train will go at 140.625 m before it is brought to rest.