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A car accelerates from $6\ ms^{-1}$ to $16\ ms^{-1}$ in $10\ sec.$ Calculate the distance covered by the car in that time.
Here initial velocity $u=6ms^{-1}$
Final velocity $v=16\ ms^{-1}$
Time $t=10\ sec.$
Therefore, acceleration $a=\frac{v-u}{t}$
$=\frac{16-6}{10}$
$=1\ m/s^2$
Let $s$ be the distance covered by the car,
On using the equation $v^2=u^2+2as$
$16^2=6^2+2\times1\times s$
Or $256=36+2s$
Or $2s=256-36$
Or $2s=220$
Or $s=\frac{220}{2}$
Or $s=110\ m$
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