Question

A train travels a distance of 60 km at uniform speed. If the speed of the train was reduced by 10 kmh-1, the time taken to travel the 60km will increase by 1/2h. Find the speed of the train at the beginning.

Answer 1

Answer:

Initial speed is 32 m/s

At uniform speed, acceleration is 0, (a = 0).

When speed reduced, (v - u) = 2.78 ms-¹, t = 1800 sec, s = 60 ,000 metres.

From first equation of motion:

${ \boxed{ \bf{v = u + at}}} \\ { \tt{(v - u) = at}}$

substitute:

${ \tt{2.78 = (a \times 1800)}} \\ { \tt{acceleration = 0.0015 \: {ms}^{ - 2} }}$

from second equation of motion:

${ \boxed{ \bf{s = ut + \frac{1}{2} a {t}^{2} }}}$

substitute:

${ \tt{60000 = 1800u + ( \frac{1}{2} \times 0.0015 \times {1800}^{2}) }} \\ { \tt{1800u = 57570}} \\ { \tt{u = 32 \: m {s}^{ - 1} }}$