Question

A train is running at 60km per hour has its speed increased under mean acceleration of 0.2m/s for 60second.what is its speed at the end of this period. How many kms will it cover during this period.

Answer 1

Intial Velocity (u)=60 km/h

In m/s=16.66 m/s=17 m/s

Accerlation(a)=0.2 m/s

Time(t)=60 Seconds

Final Velocity (v)=??

Use Motion of Equation

v=u+at

v=17+0.2×60

v=17+12

v=29 m/s

Final Velocity=29 m/s

Intial Velocity=17 m/a

Time=60 seconds

Accerlation=0.2 m/s

Distance (s)=??

Use Equation of Motion.

v²=u²+2as

29²=17²+2×0.2×s

29²-17²=0.4s

0.4s=29²-17²

0.4s=(29-17)(29+17)

0.4s=12×46

0.4s=552

s=552÷0.4

s=1380 m

In km=1.38 km

$\boxed{\mathbf{Speed=29 m/s}}$

$\boxed{\mathbf{Distance=1.38 km}}$