Question

the driver of a car travelling at 72km/h applies the break and acceleration uniformly in the opposite direction at the rate of 5m/s .how long the car will take to come to rest ?how far will the car cover before coming to rest​

Answer 1

Given :

Initial velocity ,u= 72km/hr=20 m/s

Final velocity , v =0 m/s [ Brakes applied ]

Accelaration , a= -5 m/s²

To Find :

• Time taken by the car to come to rest
• Distance covered by the car before coming to rest

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

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

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

Part -1

We have to Find Time taken by the car to come to rest

By First Equation of Motion

$\rm\:v=u+at$

Put the given values

$\sf\implies\:0=20-5\times\:t$

$\sf\implies\:5t=20$

$\sf\implies\:t=\dfrac{20}{5}=4sec$

Therefore , Time taken by the car to come to rest is 4 sec

Part -2

We have to find Distance covered by the car before coming to rest

By using third Equation of motion

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

Put the given values , then

$\sf\implies\:0=(20)^2+2(-5)S$

$\sf\implies\:10S=400$

$\sf\implies\:S=\dfrac{400}{10}$

$\sf\implies\:S=40m$

Therefore, Distance covered by the car before coming to rest is 40 m.

Answer 2

❍$\Large{\underline{\sf{Step\:by\:step\:explanation:}}}$

$\bf Given \begin{cases} & \sf{Initial\:velocity, \sf{u=72km/hr=20m/s}} \\ & \sf{Final\:velocity , \sf{v=0\:m/s\:\big[ Brakes\:applied \big]}} \\ & \sf{Acceleration , \sf{a=-5\:m/s^{2}}} \end{cases}$

We need to find the time the car will take to come to rest and the distance the car will cover before coming to rest.

• As the car decelerates at the rate of 5m/s.

So, the deceleration is, $\displaystyle{\sf{a=\: -5m/s^{2}}}$

• So as the 1st equation of motion states ★$\displaystyle{\boxed{\sf{\orange{v=u+at}}}}$★

Substituting the values, we get

And v = 0, as the car ultimately comes to rest.

$\implies{\sf{0=20-5t}}$

$\large\implies{\boxed{\sf{\red{t=4\:secs}}}}$

Time taken by the car to stop is 4s.

_____________________

We know that, the 2nd equation of motion states

• ★$\displaystyle{\boxed{\sf{\orange{v^{2}=u^{2}+2as}}}}$★

Substituting the values, we get

$\implies{\sf{0=400+2 \times 5 \times s}}$

$\large\implies{\boxed{\sf{\red{s=40m}}}}$

Distance covered by the car to stop is 40m.

__________________________