First slide

The 3 equation for uniform acc

Question

A train normally travels at a uniformly speed of 72 km/h on a long stretch of straight level track. On a particular day, the train was forced to make a 2.0 minute stop at a station on this track. If the train decelerates at a uniform rate of 1.0 $m / s^{2}$ and accelerates at a rate of 0.50 $m / s^{2}$ , how much time is lost in stopping at the station?

Difficult

A

2 min
B

2 min 30 s
C

30 s
D

2 min 20 s

Solution

Case -I: Actual time spent in decelerating
t= v-u/a=20s
Distance travelled in decelerating
= $20 \times 20 - \frac{1}{2} \times 1 \times 20^{2} = 400 - 200 = 200 m$
Time that train would have taken if it had travelled uniformly with 20 m/s for 200 m = $\frac{200}{20} = 10 s$
Extra time spent in decelerating = 20-10 = 10 s
Case-II : Actual time spent in accelerating
t= $\frac{v - u}{a} = \frac{20 - 0}{0.5} = 40 s$
Distance travelled in these 40 s
= $0 \times 40 + \frac{1}{2} \times 0.5 \times 40^{2} = 400 m$
Time that the train would have taken if it had travelled uniformly with 20 m/s = $\frac{400}{20} = 20 s$
Extra time lost due to acceleration = 40-20 = 20 s
Total extra time = 10s + 2 min + 20 s = 2 min 30 s

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