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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Updated on: 10-Oct-2022

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