A train accelerating uniformly from rest attains a maximum speed of 40 ms−1 in 20 s. It travels at this speed for 20 s and is brought to rest under uniform retardation in further 40 s. What is the average velocity during this period?
A
80/3ms−1
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B
40ms−1
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C
25ms−1
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D
30ms−1
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Solution
The correct option is C25ms−1 ∴40=(20)a1 ∴a1=2m/s2
Further 40=(40)a2 ∴a2=1m/s2
Therefore acceleration is 2m/s2 and retardation 1m/s2
Now, S1=12a1t21=12×2×(20)2=400m S2=vmaxt2=40×20=800m S3=V2max2a2=(40)22×1=800m
Now average velocity =Total displacementTotal time =400+800+80020+20+40=25m/s.