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(D) The work done (energy required) to change the kinetic energy of an object is given by the work-energy theorem: $W = \Delta K.E. = \frac{1}{2}m(v_f

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(D) The work done (energy required) to change the kinetic energy of an object is given by the work-energy theorem: $W = \Delta K.E. = \frac{1}{2}m(v_f^2 - v_i^2)$.For the first case,accelerating from $10 \, m/s$ to $20 \, m/s$:$W_1 = \frac{1}{2}m(20^2 - 10^2) = \frac{1}{2}m(400 - 100) = \frac{1}{2}m(300) = 150m$.For the second case,accelerating from rest $(0 \, m/s)$ to $10 \, m/s$:$W_2 = \frac{1}{2}m(10^2 - 0^2) = \frac{1}{2}m(100 - 0) = 50m$.Comparing the two energies:Ratio $= \frac{W_1}{W_2} = \frac{150m}{50m} = 3$.Therefore,the energy required is $3$ times the energy required to accelerate from rest to $10 \, m/s$.

The energy required to accelerate a car from $10 \, m/s$ to $20 \, m/s$ is how many times the energy required to accelerate the car from rest to $10 \, m/s$?

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