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Q. The velocity of a particle at an instant is $10\, m\, s^{-1}$. After $3\, s$ its velocity will become $16 \,m\, s^{-1}$. The velocity at $2 \,s$, before the given instant will be

Motion in a Straight Line

Solution:

Here, $u = 10 \,m \,s^{-1}$, $t = 3 \,s$, $v = 16 \,m \,s^{-1}$
$a=\frac{v-u}{t}=\frac{16-10}{3}=2\,m\,s^{-2}$
Now velocity at $2\, s$, before the given instant
$10=u+2\times 2$
$\therefore u=6\,m\,s^{-1}\quad\left(\because v=u+at\right)$