Question

A car initially at rest, accelerates at 7.29 m/s to the 2nd power for 11.3 s. How far did it go in this time?

Answer 1

To be given in question

Car initial velocity $u=0$

$v=7.29 meter/sec$

$t=11.3sec$

To be asked in question

Distance $s$ =?

First law of newton motion

$v=u+at$

Put value

$7.29=0+a\times11.3$

a=0.645$meter/sec^2$

second law of newton motion

$s=ut+\frac{1}{2}at^2$

$=0+\frac{1}{2}\times0.645\times(11.3)^2$

$s=41.19 meter$