Question

3) An object moving with a speed of 50 m/s takes 8 second to come to rest. The distance

Travelled during this time duration is -

Answer 1

Given :

Initial velocity ,u= ,50 m/s

Final velocity , v = 0

Time , t = 8 sec

To Find :

The distance travelled by the object during this time duration

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion.

$\bf\:v=u+at$

$\bf\:S=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2aS$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

${\underline{\sf{Answer}}}$

We have to the distance travelled by tje object during this time duration.

By First Equation of motion

$\rm\:v=u+at$

Put the given values

$\sf\implies\:0=50+a\:\times8$

$\sf\implies\:8a=-50$

$\sf\implies\:a=\dfrac{-50}{8}$

$\sf\implies\:a=-6.25ms^{-2}$

Now ,

By Third Equation of motion

$\rm\:v{}^{2}=u{}^{2}+2aS$

Put the given values

$\sf\implies\:0=(50)^2+2\times(-6.25)\:S$

$\sf\implies\:12.5S=2500$

$\sf\implies\:S=\dfrac{2500}{12.5}$

$\sf\implies\:S=\dfrac{25000}{125}$

$\sf\implies\:S=200m$

Therefore, the distance travelled by the object during this time interval is 200m .