Question

a train is travelling at a speed of 54 km/h^-1 brakes are applied so as to produce a uniform acceleration of 0.25 ms^-2.calculate the distance travelled by the train before it comes to rest

Answer 1

Explanation:

Given that,

Acceleration a=−0.5m/s

2

Speed v=90km/h=25m/s

Using equation of motion,

v=u+at

Where,

v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the equation

Finally train will be rest so, final velocity,v=0

0=25−0.5t

25=0.5t

t=

0.5

25

t=50 sec

Again, using equation of motion,

S=ut+

2

1

at

2

Where, s = distance

v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the equation

Where S is distance travelled before stop

s=25×50−

2

1

×0.5×(50)

2

s=625 m

So, the train will go before it is brought to rest is 625 m.

Hence, A is correct

Answer 2

Answer:

here a = -0.5 ms^-2

final velocity Vf = 0

initial velocity Vi = 54 km^-h = 15 ms^-1

s=?

now we know:

Vf^2 = Vi^2 + 2as

=> 2as = - Vi^2

=> s = -Vi^2/(2a)

=> s = -225/(2*(-0.5))

=> s = 225m