Question

A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a final velocity v then t is increased so that the displacement is 3x. What final velocity does the motorcycle attain?

Answer 1

As a very first step, one could read the question properly and hence note down the given quantities by dividing them into two parts as provided. Then you could accordingly choose the equation of motion and hence apply it for the two parts and then relate both of them to get the answer.

Formulas used:
Equation of motion,
${{v}_{f}}^{2}-{{v}_{i}}^{2}=2as$

Complete step-by-step solution:
In the question, we are given two sets of conditions. First set of condition is as follows:
Initial velocity of the motorcycle would be,
${{v}_{i}}=0$since it said to start from rest.
The final velocity of the motorcycle is given to be,
${{v}_{f}}=v$
Let the constant acceleration for the given motorcycle be a and its displacement after t seconds is given to be x.
Here the equation of would be,
${{v}_{f}}^{2}-{{v}_{i}}^{2}=2as$
$\Rightarrow {{v}^{2}}-0=2ax$……………………………………… (1)
Now for the second part we have the displacement to be 3x and we are supposed to find the final velocity for the same.
Here the equation of motion would be,
$v{{'}^{2}}-0=2a\left( 3x \right)=3\left( 2ax \right)$
Substituting equation (1) here, we get,
$v{{'}^{2}}=3{{v}^{2}}$
$\therefore v'=3v$
Therefore, we found the velocity of the motorcycle for the second part to be $\sqrt{3}$times the final velocity in the first part.
Note: The gravitational force is equal and opposite on the two bodies involved but the force on Earth due to objects on Earth is negligible due to difference in mass. Any change in acceleration due to gravity near the Earth’s surface is negligible. The Coulomb’s law of gravitation follows the inverse square law.