Question

A car accelerates from rest to\[30\;{\rm{km/h}}\]. Later, on a highway it accelerates from\[30\;{\rm{km/h}}\] to\[60\;{\rm{km/h}}\]. Which takes more energy, going from 0 to 30, or from 30 to 60?

1. 0 to\[30\;{\rm{km/h}}\] .
2. 30 to\[60\;{\rm{km/h}}\].
3. Both are the same.

Short answer

The correct answer is option (b): 30 to \(60\;{\rm{km/h}}\).

Step-by-step solution

Definition of kinetic energy

The energy possessed by an object by virtue of its motion is called kinetic energy.For example, the motion of a yo-yo, stretching a rubber band, moving the muscles, a moving car, etc.

Formula of kinetic energy

The kinetic energy is proportional to the mass and square of the velocity of the object. Its formula is given by:

\(KE = \frac{1}{2}m{v^2}\)

Here, m is the mass of the object, and vis the velocity.

Determination of the kinetic energy of the car when going from 0 to \(30\;{\rm{km/h}}\)

The change in the kinetic energy is given by:

\(KE = \frac{1}{2}m\left( {v_2^2 - v_1^2} \right)\)

Here,\({v_1} = 0\)and\({v_2} = 30\;{\rm{km/h}}\).

Therefore, the change in energy will be:

\(\begin{aligned}KE &= \frac{1}{2}m\left( {{{\left( {30\;{\rm{km/h}}} \right)}^2} - {{\left( 0 \right)}^2}} \right)\\ &= \frac{1}{2}m\left( {900} \right)\\ &= 450m\end{aligned}\)

Determination of the kinetic energy of the car when going from \(30\;{\rm{km/h}}\) to \(60\;{\rm{km/h}}\)

The change in the kinetic energy is given by:

\(KE = \frac{1}{2}m\left( {v_2^2 - v_1^2} \right)\)

Here,\({v_1} = 30\;{\rm{km/h}}\)and\({v_2} = 60\;{\rm{km/h}}\).

Therefore, the change in energy will be:

\(\begin{aligned}KE &= \frac{1}{2}m\left( {{{\left( {60\;{\rm{km/h}}} \right)}^2} - {{\left( {30\;{\rm{km/h}}} \right)}^2}} \right)\\ &= \frac{1}{2}m\left( {3600 - 900} \right)\\ &= 1350m\end{aligned}\)

Thus, it takes three times more energy to go from 30 to \(60\;{\rm{km/h}}\).