Question

A particle moves along a straight line. The acceleration of the particle as function of time is given by $a=6t–12$. Initial velocity of the particle is $9m/s$ and the distance travelled by the particle in $5s$ is $7x$(in $m$). Find the value of $x$.

Answer 1

Given,
$a=6t-12u=9m/s$
Distance travelled in $5s=7x$
As we know.
$a=\frac{dv}{dt}...\left(i\right)adt=dv$
Integrating both sides by putting the values we get,
${\int }_9^v=dv={\int }_0^t(6t-12)dtv-9=3t^2-12tv=3t^2-12t+9=0$
at $t=1s$ and $t=3s$

As we know,
$v=\frac{ds}{dt}...\left(ii\right)\frac{ds}{dt}={\int }_3^{5}3t^2-12t+9{\int }_0^{s}ds={\int }_3^5(3t^2-12t+9)dt$

Hence,
$s=|\int (3t^2-12t+9{)}_0^{1}dt|+|\int (3t^2-12t+9{)}_1^{3}dt|+|\int (3t^2-12t+9{)}_3^5|dts=28m$

Therefore,
$7x=28x=4m$