Question

A car moving with a speed of 40 km/h can be stopped by applying brakes at least after 2 m. If the same car is moving with a speed of 80km/h , what is the minimum stopping distance?

Answer 1

Answer:

8m

Explanation:

as stopping vel.or final is 0

so.,v^2=u^2-2as

we can conclude that,

u^2=2as

u^2 directly proportional to s

Answer 2

The minimum stopping distance of the car is 8 m

Explanation:

Let the acceleration of the car is a

Initial speed u =40 km/h

Final speed v = 0

Stopping distance s = 2 m = 0.002 km

Therefore, by using the third equation of motion

$v^2=u^2-2as$

$0^2=40^2-2\times a\times 0.002$

$\implies a=\frac{40\times 40}{2\times 2\times 10^{-3}}$

$\implies a=4\times 10^5$ km/h²

Now if the initial velocity u' = 80 km/h

Then, again from the third equation of motion

Stopping distance s'

$s'=\frac{u'^2}{2a}$

$\implies s'=\frac{80\times 80}{2\times 4\times 10^5}$

$\implies s'=8\times 10^{-3}$ km

or, $s'=8$ m

Therefore, the stopping distance will be 8 m

Hope this answer is helpful.

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