Question

If the speed of a vehicle become 3 times for a given deceleration its stopping distance

become ......

(A) 2 times
(B) 9 times
(C)
19
times
(D) 3 times​

Answer 1

Answer:

B 9 times

d is proportional to velocity square

v^2=2ad

Answer 2

Answer:

The stopping distance increases by 9 times.

Explanation:

Given a condition, the speed of a vehicle become 3 times for a given deceleration

Let the speed of the vehicle be u m/s.

And let the deceleration be $a~m/s^{2}$.

The stopping distance of a vehicle is the distance that a vehicle travels from the moment, the brakes are applied to the vehicle stops.

Mathematically given by a simple formula, $d=-\frac{u^{2} }{2a}$

where the negative sign shows that the acceleration is in the opposite direction to the motion of the vehicle i.e., showing deceleration.

Ignoring the negative sign, the stopping distance is

$d_1=\frac{u^{2} }{2a}$

Given that speed is increased 3 times, therefore the new stopping distance is

$d_2=\frac{(3u)^{2} }{2a}=\frac{ 9u^{2} }{2a}$

Acceleration is same in both cases, hence the change in the stopping distance is

$\frac{d_2}{d_1} =\frac{\frac{ 9u^{2} }{2a}}{\frac{u^{2} }{2a}}$

$=\frac{ 9u^{2} }{2a}\times\frac{2a }{u^{2}}=9$

Hence, the stopping distance increases by 9 times.

So, the correct answer is option 2.