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Time taken by a train to slow down from 80 kmh-1 to 20 kmh-1 with a uniform deceleration of 2 ms-2 is
- a)9 s
- b)10 s
- c)8.3 s
- d)8.4 s
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Time taken by a train to slow down from 80 kmh-1to 20 kmh-1with a unif...
Given:
Initial velocity, u = 80 km/h = 22.22 m/s
Final velocity, v = 20 km/h = 5.56 m/s
Acceleration, a = -2 m/s^2 (negative sign indicates deceleration)
We need to find the time taken by the train to slow down from 80 km/h to 20 km/h with a uniform deceleration of 2 m/s^2.
Using the formula v = u + at, we can find the time taken as follows:
v = u + at
=> t = (v - u) / a
=> t = (5.56 - 22.22) / (-2)
=> t = 8.33 s (approx.)
Therefore, the time taken by the train to slow down from 80 km/h to 20 km/h with a uniform deceleration of 2 m/s^2 is 8.33 s (approx.), which is closest to option (c) 8.3 s.
Initial velocity, u = 80 km/h = 22.22 m/s
Final velocity, v = 20 km/h = 5.56 m/s
Acceleration, a = -2 m/s^2 (negative sign indicates deceleration)
We need to find the time taken by the train to slow down from 80 km/h to 20 km/h with a uniform deceleration of 2 m/s^2.
Using the formula v = u + at, we can find the time taken as follows:
v = u + at
=> t = (v - u) / a
=> t = (5.56 - 22.22) / (-2)
=> t = 8.33 s (approx.)
Therefore, the time taken by the train to slow down from 80 km/h to 20 km/h with a uniform deceleration of 2 m/s^2 is 8.33 s (approx.), which is closest to option (c) 8.3 s.
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Community Answer
Time taken by a train to slow down from 80 kmh-1to 20 kmh-1with a unif...
v= 80*1000/3600 = 22.22
u=20*1000/3600=5.55
Now a = v-u/t
= 22.22-5.55/2
= 8.335
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