
Stopping distance of a moving vehicle is directly proportional to
A. Square of initial velocity
B. Square of the initial acceleration
C. The initial velocity
D. The initial acceleration
Answer
541.5k+ views
Hint: To solve this question we have to know what initial velocity is, what is final velocity and what is acceleration. We know that the velocity is the speed for which any object can move in a particular direction. So when the object starts moving then that velocity is called initial velocity.
Complete step by step answer:
We can assume that the initial velocity of a moving vehicle is u.According to the question, we have to find the stopping distance that means that vehicle will stop after some time. So, we can say that the final velocity will be zero. We can call final velocity as v. so here, v is equal to zero. And we can assume acceleration is a. so, here will be negative. Which means here it is not acceleration. It is retardation. We are now going to apply third equation of motion which is,
\[{v^2} = {u^2} + 2as\]
Now, after putting the value of v we will get,
$0 = {u^2} + 2( - a)s \\
\Rightarrow - {u^2} = - 2as \\
\Rightarrow s = \dfrac{{{u^2}}}{{2a}} \\$
So, here we can clearly see that \[2a\] is a constant. So, \[s \propto {u^2}\].We can say that the stopping distance of a moving vehicle is directly proportional to the square of initial velocity. Because we assumed that u is the initial velocity.
Hence, option A is the right answer.
Note: We can get confused between the sign of acceleration. We have to keep it in our mind that when any object is going to stop then the acceleration will be negative which is called retardation. Otherwise we can make mistakes while doing these types of questions.
Complete step by step answer:
We can assume that the initial velocity of a moving vehicle is u.According to the question, we have to find the stopping distance that means that vehicle will stop after some time. So, we can say that the final velocity will be zero. We can call final velocity as v. so here, v is equal to zero. And we can assume acceleration is a. so, here will be negative. Which means here it is not acceleration. It is retardation. We are now going to apply third equation of motion which is,
\[{v^2} = {u^2} + 2as\]
Now, after putting the value of v we will get,
$0 = {u^2} + 2( - a)s \\
\Rightarrow - {u^2} = - 2as \\
\Rightarrow s = \dfrac{{{u^2}}}{{2a}} \\$
So, here we can clearly see that \[2a\] is a constant. So, \[s \propto {u^2}\].We can say that the stopping distance of a moving vehicle is directly proportional to the square of initial velocity. Because we assumed that u is the initial velocity.
Hence, option A is the right answer.
Note: We can get confused between the sign of acceleration. We have to keep it in our mind that when any object is going to stop then the acceleration will be negative which is called retardation. Otherwise we can make mistakes while doing these types of questions.
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