Question

Stopping distance of a moving vehicle is directly proportional to

• A. square of the initial velocity
• B. square of the initial acceleration
• C. the initial velocity
• D. the initial acceleration

Answer 1

The Correct Option is A

Let $d_s$ is the distance travelled by the vehicle before it stops. Here, final velocity $v = 0$, initial velocity $= u$, $S = d_s$ Using equation of motion $v^2=u^2+2aS$ $\therefore \left(0\right)^{2}=u^{2}+2ad_{s}$; $d_{s}=-\frac{u^{2}}{2a}$ or $d_{s} \propto u^{2}$

Answer 2

We can suppose that a moving vehicle's starting velocity is u. According to the question, we must determine the stopping distance, which indicates that the car will come to a halt after a certain amount of time. As a result, the final velocity will be zero. The final velocity is denoted by the symbol v. As a result, v is equal to zero in this case. We might also suppose that acceleration is a. So It'll be negative here. That is, there is no acceleration here. It's called retardation. We will now apply the third equation of motion, which is,

v²=u²+2as
Now, after putting the value of v we will get,
0=u²+2(−a)s
⇒ −u²=−2as
⇒ s=\(\frac{u^2}{2a}\)
So, here we can clearly see that 2a is a constant. So, S∝u².
A moving vehicle's stopping distance is exactly related to the square of its beginning velocity. We assumed that u represents the starting velocity.