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Q.
An object accelerates from rest to a velocity $27.5\,m/s $ in $10\,\sec $ then the distance covered in next $ 10\,\sec$ is
Punjab PMETPunjab PMET 2005Motion in a Straight Line
Solution:
Given : $u=0,\, v=27.5\, m / s,\, t=10\, \sec$
From first equation of metion
$v =u+ a t$
$27.5 =0+a \times 10$
$a =2.75 m / s ^{2}$
Distance covered in first $10\, \sec$ is
$s_{1}=u t+\frac{1}{2} a t^{2}$
$=0 \times 10+\frac{1}{2} \times 2.75 \times(10)^{2}$
$=\frac{1}{2} \times 2.75 \times 100$
$=137.5\, m$
Distance covered in next $10\, \sec$
with uniform velocity of $27.5\, m / s$
$s_{2}=27.5 \times 10=275\, m$
Total distance covered is
$=137.5+275=412.5\, m$