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


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Answer 1

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

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

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.

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 ★$\displaystyle{\boxed{\sf{\red{v=u+at}}}}$

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.

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We know that, the 2nd equation of motion states

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

Substituting the values, we get

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

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

Distance covered by the car to stop is 40m.

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Answer 2

Explanation:

u = 72km/h

= 72*5/18

= 20m/s

a= -5m/s^-2 ( negative sign because it is retardation)

v= 0m/s (as the break applies)

Equation of motion

v= u+at

therefore t= v-u/a

t= 0-20/-5

t= 4s

by second equation of motion

s= ut+1/2at^2

s= 20*4+0.5*-5*4^2

s= 80-40

s= 40m