A TRAIN MOVING ALONG A STRAIGHT TRACK, ACCELERATES FROM 36 KM/H TO 72 KM/H IN 5 SECONDS. HOW MUCH DISTANCE DOES IT COVER IN THIS TIME
Answers
Answered by
76
- 75 metres is the required answer.
Given:-
- Initial velocity ,u = 36km/h
- Final velocity ,v = 72km/h
- Time taken ,t = 5s
To Find:-
- Distance Covered ,S
Solution:-
According to the Question
Firstly we change the unit here into m/s
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Initial velocity ,u = 36
→ u = 36×5/18
→ u = 2×5
→ u = 10m/s
Again,
Final velocity ,v = 72
→ v = 72×5/18
→ v = 4×5
→ v = 20m/s
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Now, calculating the acceleration of the
- a = v-u/t
where,
- v is the final velocity
- a is the acceleration
- u is the initial velocity
- t is the time taken
Substitute the value we get
a = 20-10/5
a = 10/5
a = 2m/s²
Using 3rd equation for calculating the distance covered by train.
- v² = u² + 2as
Substitute the value we get
20² = 10² + 2×2 × s
400 = 100 + 4s
400-100 = 4s
300 = 4s
4s = 300
s = 300/4
s = 75 m
- Hence, the distance covered by the train in the given time interval is 75 metres.
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Answered by
66
Explanation :
- Distance covered (s) = ?
- Firstly, changing units of initial velocity and final velocity into m/s :
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- Now, finding it's acceleration :
- Using 3rd eqn of motion for finding distance travelled by a train :
- Three equations of motion :
- v = u + at
- s = ut + (1/2) at²
- v² = u² + 2as ㅤㅤㅤ[Used above]
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