Question

A train is travelling at a speed of 90km/h. Brakes are applied so as to

produce a uniform negative acceleration of - 0.5m/s2

. Find how far the train

will go before it is brought to rest.​

Answer 1

625m

U= 90km/h= 25m/s

V= 0

A= -0.5m/s^2

S= ?

2AS= V^2-U^2

2*-0.5*S= -25^2

S= 625m

Answer 2

Answer:

So,the distance travelled by the train is 625m.

Explanation:

train's speed=

u = 90 km/h = 90× 5/18= 25 m/s

a= -0.5m/s^2

{negative sign is used for negative acceleration}

now we need value of t ,so we apply:-

v= u+at

final velocity is zero because finally train will be at rest.

v= u+at

0= 25- 0.5t

t= 50 sec

# for distance

S = ut+ 1/2at^2

S = 25×50 + 1/2×0.5×(50)^2

S = 1250 - 1/2 × 0.5 × 2500

S = 1250-625

S = 625m