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
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
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