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# A race scooter is seen accelerating uniformly from $18.5\ m/s$ to $46.1\ m/s$ in $2.47$ seconds. Determine the acceleration of the scooter and the distance travelled.

Here, initial velocity $u=18.5\ m/s$

Final velocity $v=46.1\ m/s$

Let $a$ be the acceleration of the race scooter.

Time taken to change the velocity of the race scooter $t=2.47\ sec.$

On substituting the above values in the equation, $v=u+at$

$46.1=18.5+a\times 2.47$

$2.47a=46.1-18.5$

Or $2.47 a=27.6$

Or $a=\frac{27.6}{2.47}$

Or $a=11.2\ m/s^2$

Therefore, the acceleration of the race scooter is $11.2\ m/s^2$.

Let $s$ be the distance travelled by the race scooter.

From the equation $v^2=u^2+2as$

$s=\frac{v^2-u^2}{2a}$

Or $s=\frac{( 46.1)^2-( 18.5)^2}{2\times11.2}$

Or $s=\frac{( 46.1+18.5)(46.1-18.5)}{22.4}$ [because $a^2-b^2=( a+b)( a-b)$]

Or $s=\frac{64.6\times27.6}{22.4}$

Or $s=79.60\ m$

Thus, the distance travelled by the race scooter is $79.60\ m$.