Question

3.
A car moving with a speed of 40 km/hr can be stopped by applying brakes after at least 2m. If the
same car is moving with a speed of 80 km/hr, what is the minimum stopping distance?
1) 8 m
2) 6 m
2 4 m
4) 2 m​

Answer 1

Answer:

1] 8m

Explanation:

Initial velocity, u = 40 km/h = 40 × (5/18) = 200/18 m/s

Final velocity, v = 0

Distance travelled before coming to rest, s = 2 m

Using, v2 = u2 + 2as

=> a = -u2/(2s)

=> a = -30.864 m/s2

Again,

u = 80 km/h = 400/18 m/s

v = 0

a = -30.864m/s2

Now, v2 = u2 + 2as

=> s = -u2/(2a) = 8 m

Answer 2

The minimum stopping distance of the car is 8 m

Therefore, option (1) is correct.

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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