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Q. A train accelerating uniformly from rest attains a maximum speed of $40 \,ms ^{-1}$ in $20\, s$. It travels at the speed for $20\, s$ and is brought to rest with uniform retardation in further $40\, s$. What is the average velocity during the period?

BITSATBITSAT 2014

Solution:

$S _{1}=\left(\frac{ u + v }{2}\right) t$
$=20 \times 20=400 \,m$
$S _{2}=40 \times 20=800\, m$
$S _{3}=\left(\frac{ u + v }{2}\right) t$
$=20 \times 40=800 \,m$
Total displacement $=2000 \,m$
Total time taken $=80\, m$
$V =\frac{2000}{80}=25 \,m / s$
Total time taken $=80\, m$
$V =\frac{2000}{80}=25\, m / s$