First slide

Rectilinear Motion

Question

train normally travels at a uniform speed of 72V,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
along this track. If the train decelerates at a uniform rate of 1.0 m/s² and accelerates at a rate of 0.50 m/s², 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
$= \frac{v - u}{a} = \frac{0 - 20}{- 1.0} = 20 s$
Distance travelled in decelerating
$= 20 \times 20 - \frac{1}{2} \times 1 \times 20^{2} = 400 - 200 = 200 m$
Time that train will 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
$= \frac{v - u}{a} = \frac{20 - 0}{0.5} = 40 s$
Distance travelled in these 40 s
$= 0 \times 40 + \frac{1}{2} \times 0.5 \times 40^{2} = 400 m$
Time that the train will have taken if travelled uniformly with 20 m/s
$= \frac{400}{20} = 20 s$
$\begin{array}{l} Extra time lost due to acceleration = 40 - 20 = 20 s \\ Total extra time = 10 s + 2 \min + 20 s = 2 min 30 s \end{array}$

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