2. A train accelerates from 36 kmph to 90 kmph on covering a distance 500 m.
Calculate the acceleration of the train and the time taken to cover the distance
500m.
Answers
Answer :
›»› The Acceleration of train is 0.525 m/s²
›»› The Time taken by train is 28.57 sec
Given :
- Initial velocity of train = 36 km/hr
- Final velocity of train = 90 km/hr
- Distance covered by train = 500 m
To Calculate :
- Acceleration of train = ?
- Time taken by train = ?
Calculation :
Here in this question we have to calculate the Acceleration and Time taken by train. So, firstly we need to convert the SI unit of Initial velocity and Final velocity of train from km/hr to m/s, after that we will calculate Acceleration and Time taken by train on the basis of conditions given above.
→ Initial velocity = 36 km/hr
→ Initial velocity = 36 * 5/18
→ Initial velocity = 2 * 5
→ Initial velocity = 10 m/s
→ Final velocity = 90 km/hr
→ Final velocity = 90 * 5/18
→ Final velocity = 5 * 5
→ Final velocity = 25 m/s
Now,
From third equation of motion
→ v² = u² + 2as
→ 25² = 10² + 2 * a * 500
→ 625 = 100 + 2 * a * 500
→ 625 - 100 = 2a * 500
→ 525 = 1000a
→ a = 525/1000
→ a= 0.525
║Hence, the Acceleration of train is 0.525 m/s².║
From first equation of motion
→ v = u + at
→ 25 = 10 + 0.525 * t
→ 25 - 10 = 0.525 * t
→ 15 = 0.525t
→ t = 15/0.525
→ t = 28.57
║Hence, the time taken by train are 28.57 sec.║
Given :-
☄ Initial velocity, u = 36 km/hr
☄ Final velocity, v = 90 km/hr
☄ Distance covered, s = 500 m
To find :-
(a) Acceleration of train
(b) Time taken by train
Conversion :-
we have to convert km/h into m/s
1km/hr = 5/18m/s
36km/hr = 36 × 5/18 = 10m/s
90km/hr = 90 × 5/18 = 25m/s
Solution :-
(a) using equation of motion .i.e.,
➠ v² = u² + 2as
➠ 25² = 10² + (2)(a)(500)
➠ 625 = 100 + (1000)(a)
➠ 625 - 100 = 1000a
➠ a = 525/1000
➠ a = 0.525
thus, the Acceleration of train is 0.525 m/s².
(b) using equation of motion .i.e.,
➠ v = u + at
➠ 25 = 10 + (0.525)(t)
➠ 15 = 0.525t
➠ t = 15/0.525
➠ t = 28.57
thus, the time taken by train is 28.57 sec.
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