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Q.
An object accelerates from rest to a velocity $27.5 \,m/s$ in $10 \,s$ , then the distance covered after next $10 \,s$ is:
Bihar CECEBihar CECE 2004
Solution:
Since, the object accelerates from rest, its initial velocity $(u)$ is zero.
That is, $u=0$ From first equation of motion
$v=u+a t$
$\therefore 27.5=0+a \times 10$
or $a=2.75 \,m / s ^{2}$
Hence, distance covered in first $10 s$
$s_{1}=u t+\frac{1}{2} a t^{2}$
$=0+\frac{1}{2} \times 2.75 \times(10)^{2}$
$=137.5\, m$
Distance covered in next $10 s$ with uniform velocity of $27.5\, m / s$
$s_{2}=27.5 \times 10=275\, m$
Total distance covered
$s=137.5+275=412.5\, m$