Question

A train is travelling at a speed of 90 km h−1. Brakes are applied so as to produce a uniform acceleration of −0.5 m s−2. Find how far the train will go before it is brought to rest.

Answer 1

V=0
u=90km/h = 25m/s
v^2=u^2+2as
0=(25)2+2×-0.5s
0=625-1×s
0=625-s
s=625 m

Answer 2

Heya friend,

Here is the answer-

u= 90km/hr = 25m/s
v=0
a= -0.5m/s^2
s=?

To find 's' we will use 3rd equation of motion, you can use 2nd as well, but then you need to find time.I am doing this by 3rd law of motion.

=

2as=v^2-u^2
2× -0.5×s = (0)^2 - (25)^2
-s= 0-625
-s= -625
s=625m

So, the distance the train will cover before it is brought to rest is 625m

NOTE-

• u=initial velocity
• v= final velocity
• a=acceleration
• s= distance

Hope it helps..!!!