Question

a train is traveling at a speed of 90 km per hour brakes are applied so as to produce a uniform acceleration of 0.5 ms find how far the train will go before it is brought to rest

Answer 1

given, u=90km/h,v=0, and a=-0.5m/s^2
90km/h={(90)(1000/3600)}m/s
=25m/s
v=u+at
0=25+(-0.5)(t)
(-0.5)(t) =-25
t=-25/-0.5
t=250/5
t=50s

s=(u+v)/2 (t)
s=(25+0)/2 (50)
s=(25)(25)
s=625m
Hence, the distance travelled by train =625m

Answer 2

Explanation:

✤ Initial velocity (u) = 90 km/h

→ 90 × 5/18 = 25 m/s

✤ Acceleration (a) = - 0.5 m/s

✤ Final velocity (v) = 0

→ By using third equation of motion,

❖ $\large\fbox\red{\sf 2as = v^2 - u^2}$

→ $\sf{2(-0.5)s = (0)^2 - (25)^2}$

→ $\sf{-1 s = -625}$

→ $\large\fbox\purple{\sf s = 625\:m}$

_________________________

❖ Three equations of motion :-

1. $\textsf{v = u + at}$
2. $\sf{s = ut + \frac{1}{2} at^2}$
3. $\sf{2 as = v^2 - u^2}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬