Question

The train is moving with a velocity of 90 km per hour it is brought to stop by applying the brakes which is produce the retardation solve 0.5 m per second find the velocity after 10 seconds and the time taken by the train to come to rest

Answer 1

$\bold{velocity\: of \: train }$= 90 km

now,

we have to apply the $\bold{equation}$,

"v = u - a×t ",

v is $\bold{final\: velocity}$,

and u is $\bold{intial\: velocity}$,

a" is $\bold{retardation}$ and ,

t " is $\bold{Time}$.

☞Velocity at the time of applying break = 90 km/h  

= 90 km/h × (5/18)

= 25 m/s 

☞$\bold{velocity\: after\: 10\: s,}$

  v = 25 - (1/2)×10 =

20 m/s

 ☞$\bold{time\: to\: come\: to\: rest}$=

0 = 25-(1/2)×t  , solving for t ,  

we get  t = 50 s

$\bold{velocity}$ =$\bold{20}$ m/s

$\bold{Time}$= $\bold{50 }$s

Answer 2

Solution:-

Velocity of Train = 90 km/hr

Changing in m/s. we get,

90 km/hr × (5/18)

=> 10 × 5/2

=> 25 m/s.

Velocity After 10 Seconds-

=> V = u - (a × t)

=> V = 25 - (0.5 ×10)

=> V = 25 - 5

=> V = 20m/s

Time taken by the train to come to rest-

v = 0

=> 0 = 25 - (1/2)t

=> 0 = ( 50 - 1t)/2

=> t = 50 second.

Hence,

Velocity = 20m/s

& Time Taken = 50 seconds.