Question

A car moving with a speed of 50 m/s can be stopped by brakes after a distance 6m. If the same car is
moving at a speed of 100 m/s, the minimum stopping distance is
(1) 6m
(2) 12 m
(3) 18 m
(4) 24 m​

Answer 1

Answer:

(4) 24m

Explanation:

Given

Speed of car= 50 km/hr

when brakes applied final velocity become zero,

So we apply third equation of motion

v²- u² = 2 as

where u  = Initial value = 50 km/hr

           v  =  Final value = 0

          a = acceleration

           s = Distance = 6 m

Applying the value in formulae

v²- 50² = 2 a × 6

a = -2500 /12

To find minimum stopping distance

v² = 2 a s  + u²

100²    = 2 × (-2500/12 ) s

100 ×100 × 6 / 2500 = s

s = 24m

Hope it helps! Have a great day!

Answer 2

Well let us have the step by step solution and the accurate solution.

FIRST PART :

u = 50 m/s

v = 0 m/s

S = 6 m

Now, v²-u² = 2 as

Or, a = - (2500/12)

Here the negative sign represents decrease in velocity that is retardation. Hence, retardation occurred is (2500/12) metre per square second.

SECOND PART :

U = 100 m/s

V = 0 m/s

Now, V²-U² = 2 aS

Or, S = (10000×12)/5000

Or, S = 24

Thus, this time distance becomes 24 metres when it is moving with an initial speed of 100 m/s.

This indicates that option (4) is your correct option.