Question

Tripling the speed of a motor car multiplies the distance needed for stopping it by
A. $3$
B. $6$
C. $9$
D. Some other number

Answer 1

Hint: Speed or velocity means a change in the distance to the change in time. If the velocity is very high, the traveling distance is very high and the time is very low. There is a relation between velocity, distance, acceleration, and time. Use the relation between velocity and distance to find the answer.

Formula used:
\[{V^2} = {U^2} + 2aS\]
Here \[V\] – Final velocity
\[U\] – Initial velocity,
\[a\] – Acceleration
\[S\] – Distance

Complete step by step solution:
Here the final velocity is zero, means \[V = 0\],
\[0 = {U^2} + 2as\]
\[{U^2} = 2as\]
If the initial velocity is increased by \[3\], So \[U = 3\]. If we substitute U=3 means, then the square of the \[U\] is \[9\]. Then only the equation is balanced.

Additional Information:
Velocity is measured in the amount of distance traveled by an object in a given amount of time. Or we can say simply, velocity equals distance traveled divided by the time it takes to get there. From the equation, if time is increased, the velocity goes down. For example, the tortoise has very low velocity, it takes more time to cover the distance. The unit of velocity is \[m{s^{ - 1}}\](meter per second). This relation was derived from the mean speed theorem or Merton rule. The mean speed theorem was published in the year \[1335\] at Merton College, Oxford by the Philosopher, Logician, Mathematician, and Calculator Willian Heytesbury.
Hence option C is correct.

Note:
Similarly, we have other equations which describe the relationship between velocity, distance, time, and acceleration. Those relationships are,
\[v = {v_0} + at\]
\[s = {s_0} + {v_0}t + \dfrac{1}{2}a{t^2}\]