Question

4 A train starting from rest attains a velocity of 72km/h in 5 minutes .assuming the acceleration is Uniform ,find a) The acceleration b) The distance travelled by train.

Answer 1

→I m giving ans in SI Units

Initial velocity=0 mps
Final velocity=72 kmph=20 mps

Time taken=5min=300sec

Using the formula v-u=at
20=300a
a=1/15 mp(s)^2

Answer 2

Initial velocity(u) = 0 m/s
Final velocity (v) = 72 km/hr 
Time (t) = 5 min  or 1/12 hours

We know,
Equation of motion:

v = u + at

Then,
a = (v-u)/t
a =72-0/1/12
a =72*12
a =864km/h² 

Distance travelled:

Equation of motion:

s = ut + 1/2at²
s = 0x1/12 +1/2(864)(1/12)²
s = (1/2)x864x(1/144)
s = 3km

∴ Acceleration = 864 km/h² ← (Answer)
∴ Distance = 3km  ← (Answer)

Good Studies!