Answer :
Answer:
Approximately [tex]5.39 \times 10^{5}\; {\rm J}[/tex] to reach [tex]25\; {\rm m\cdot s^{-1}}[/tex].
An additional [tex]1.62 \times 10^{6}\; {\rm J}[/tex] would be required to accelerate from [tex]25\; {\rm m\cdot s^{-1}}[/tex] to [tex]50\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
When an object of mass [tex]m[/tex] travels at a speed of [tex]v[/tex], the kinetic energy of that object would be:
[tex]\displaystyle (\text{KE}) = \frac{1}{2}\, m\, v^{2}[/tex].
Initially, when the vehicle is at rest, speed would be [tex]0\; {\rm m\cdot s^{-1}}[/tex] and kinetic energy would also be [tex]0\; {\rm J}[/tex]. At [tex]v = 25\; {\rm m\cdot s^{-1}}[/tex], the kinetic energy of this [tex]m = 1\, 725\; {\rm kg}[/tex] vehicle would be:
[tex]\begin{aligned} (\text{KE}) &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2}\, (1\, 725\; {\rm kg})\, (25\; {\rm m\cdot s^{-1}})^{2} \\ &= 539\, 062.5\; {\rm J} \\ &\approx 5.39 \times 10^{5}\; {\rm J}\end{aligned}[/tex].
At twice that speed ([tex]50\; {\rm m\cdot s^{-1}}[/tex]), the kinetic energy of the vehicle would be:
[tex]\begin{aligned} (\text{KE}) &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2}\, (1\, 725\; {\rm kg})\, (50\; {\rm m\cdot s^{-1}})^{2} \\ &= 2\, 156\, 250\; {\rm J} \\ &\approx2.16 \times 10^{6}\; {\rm J}\end{aligned}[/tex].
The energy difference would be approximately:
[tex](2.16 \times 10^{6}\; {\rm J}) - (5.39\times 10^{5}\; {\rm J}) \approx 1.62 \times 10^{6}\; {\rm J}[/tex].
Because kinetic energy is proportional to the square of velocity, the energy required to double the velocity from [tex]v[/tex] to [tex]2\, v[/tex] is three times ([tex]2^{2} - 1^{2} = 3[/tex]) the energy required to reach a velocity of [tex]v[/tex] from rest.
