The chemical potential energy stored in a battery is converted into kinetic energy in a toy car that increases its speed first from $0 \text{ mph}$ to $2 \text{ mph}$ and then from $2 \text{ mph}$ up to $4 \text{ mph}$. Ignore the energy transferred to thermal energy due to friction and air resistance. Compared to the energy required to go from $0$ to $2 \text{ mph}$, the energy required to go from $2$ to $4 \text{ mph}$ is

$ \textbf{(A)}\ \text{half the amount.}$

$ \textbf{(B)}\ \text{the same amount.}$

$ \textbf{(C)}\ \text{twice the amount.}$

$ \textbf{(D)}\ \text{three times the amount.}$

$ \textbf{(E)}\ \text{four times the amount.} $
1. **Identify the formula for kinetic energy:**
   The kinetic energy \( KE \) of an object with mass \( m \) and velocity \( v \) is given by:
   \[
   KE = \frac{1}{2}mv^2
   \]

2. **Calculate the kinetic energy at different speeds:**
   - At \( v = 0 \) mph:
     \[
     KE_0 = \frac{1}{2}m(0)^2 = 0
     \]
   - At \( v = 2 \) mph:
     \[
     KE_2 = \frac{1}{2}m(2)^2 = 2m
     \]
   - At \( v = 4 \) mph:
     \[
     KE_4 = \frac{1}{2}m(4)^2 = 8m
     \]

3. **Determine the energy required for each speed increase:**
   - Energy required to go from \( 0 \) to \( 2 \) mph:
     \[
     \Delta KE_{0 \to 2} = KE_2 - KE_0 = 2m - 0 = 2m
     \]
   - Energy required to go from \( 2 \) to \( 4 \) mph:
     \[
     \Delta KE_{2 \to 4} = KE_4 - KE_2 = 8m - 2m = 6m
     \]

4. **Compare the energy required for each speed increase:**
   - The ratio of the energy required to go from \( 2 \) to \( 4 \) mph to the energy required to go from \( 0 \) to \( 2 \) mph is:
     \[
     \frac{\Delta KE_{2 \to 4}}{\Delta KE_{0 \to 2}} = \frac{6m}{2m} = 3
     \]

Thus, the energy required to go from \( 2 \) to \( 4 \) mph is three times the amount of energy required to go from \( 0 \) to \( 2 \) mph.

\(\blacksquare\)

The final answer is \( \boxed{ \text{three times the amount} } \)