Physics, asked by ahujavansh101, 8 months ago

A train is moving with a velocity of 90 kmh-1

.It is brought to stop by applying the brakes

which produces a retardation of 0.5 ms-2


. Find:


a) The velocity after 10 s.


b) The time taken by train to come to rest.

Answers

Answered by maly9570
8

Answer:

Equation to be applied  " v = u - a×t ", v is final velocity, u is initial velocity, a is retardation and t is time.

 

Velocity at the time of applying break = 90 km/h  = 90 km/h × (5/18) = 25 m/s  

 

(1) velocity after 10 s :-   v = 25-(1/2)×10 = 20 m/s

 

(2) time to come to rest :-  0 = 25-(1/2)×t  , solving for t ,  we get  t = 50 s

Answered by Anonymous
49

Given

A train is moving with a velocity of 90 kmh-1

.It is brought to stop by applying the brakes

which produces a retardation of 0.5 ms-2

To find

a) The velocity after 10 s.

b) The time taken by train to come to rest.

Solution

  • Initial velocity = 90km/h
  • Acceleration = -0.5 m/s²
  • Velocity after 10s = ?

→ u = 90m/s

→ u = 90 × 5/18

→ u = 25m/s

According to the first equation of motion

→ v = u + at

→ v = 25 + (-0.5) × 10

→ v = 25 - 5

→ v = 20m/s

Velocity after 10s is 20m/s

  • Total taken to come in rest = ?
  • Final velocity = 0 {brakes applied}

→ v = u + at

→ 0 = 25 + (-0.5) × t

→ - 25 = -0.5t

→ t = 25/0.5 = 50s

Time taken to come in rest is 50s

Note : (-) minus shows retardation

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