A train starting from rest attains a velocity of 72 km/h in 5 minutes assuming that the acceleration is uniform find :-
(I) the acceleration and
(ii) the distance travelled by the train for attaining this velocity
Answers
[1]
GIVEN :-
- Initial velocity , u = 0 m/s. [ Train starting from rest ]
- Final velocity , v = 72 km/h.
- Time , t = 5 minutes.
TO FIND :-
- The acceleration.
SOLUTION :-
we will convert the final velocity into m/s and time in seconds.
→ Final velocity , v = (72 × 1000)/3600 m/s
→ Final velocity , v = 20 m/s.
As we know that, 1 minutes = 60 seconds.
→ Time , t = 60 × 5
→ Time , t = 300 seconds.
Now as we know that,
→ a = (v - u)/t
- a = Acceleration .
- v = final velocity.
- u = initial velocity.
- t = time taken.
→ (20 m/s - 0 m/s)/300 s
→ (20 m/s)/300 s
→ (2 m/s)/30 s
→ Acceleration = 1/15 m/s².
Hence the required acceleration is 1/15 m/s².
[2]
GIVEN :-
- Initial velocity , u = 0 m/s. [ Train starting from rest ]
- Final velocity , v = 72 km/h.
- Time , t = 5 minutes.
TO FIND :-
- The distance travelled by the train for attaining the velocity.
SOLUTION :-
★ By using third equation of motion ★
→ v² - u² = 2as
- a = Acceleration .
- v = final velocity.
- u = initial velocity.
- s = Distance covered.
→ (20)² - (0)² = 2 × 1/15 × s
→ 400 - 0 = 2/15 s
→ 400 = 2/15 s
→ s = (400 × 15)/2
→ s = 200 × 15
→ s = 3000 m
Hencedistance travelled by the train for attaining the velocity is 3000 m or 3 km
Answer:
Concept Insight: - 1. When an object starts from rest, initial velocity, u = zero
2. Convert all the quantities in the same unit system and then proceed with the calculations.
u and v should be in m/s and time should be in seconds (s).
Initial velocity , u = 0m/s
Final velocity, v = 72km/h = 72 x 5/18 m/s = 20 m/s
Time, t = 5 min = 5 x 60 = 300 s
(i) Acceleration, a = ?
Acceleration, a = v- u
t
= 20 – 0 = 1 m/s2 = 0.067 m/s2
300 15
(ii) Distance, s = ?
Using 3nd equation of motion:
2as = v2 - u2
2 x 1 x s = (20)2 - (0)2
15
2 x s = 400
15
s = 400 x 15 = 3000 m = 3 km
2