Question

A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming the acceleration is uniform, find
(i) the acceleration.
(ii) the distance travelled by the train for attaining this velocity.

Answer 1

Given :

u=0m/s

V= 72km/h

T =5 min =5/60 h =1/12 h

=a=?

we know  , a=v-u/t

⇒a=72-0/1/12

=72*12

=864km/h²

s=ut+1/2at²

=0x1/12 +1/2(864)(1/12)²

=(1/2)x864x(1/144)

=3km

∴Distance travelled by the train  3km.

Answer 2

Intial Velocity of Train at Rest=0 km/h

Final Velocity of Train=72 km/h

Time (t)=5 minutes=5/60 hours

Use Equation of Motion

v=u+at

Where V=Final Velocity

U=Intial Velocity

A=Acceleration

T=Time

Put the value

72=0+5/60a

72=5/60a

a=72×60/5

a=72×12

a=864

(1)Acceleration=864 km/h²

Now

Given

Initial Velocity(u)=0 km/h

Final Velocity(v)=72 km/h

Time(t)=5/60 hours

Acceleration(a)=864 km/h²

Distance=s

Use Equation of Motion

v²=u²+2as

72²=0²+2×864×s

5184=0+1728s

5184=1728 s

s=5184/1728

s=3

Distance Travelled=3 km

$\underline{\mathbf{\huge{Acceleration=864\:km/{h}^{2}}}}$

$\underline{\mathbf{\huge{Distance=3\:km}}}$