(a)
![2.98\cdot 10^(10) J](https://img.qammunity.org/2020/formulas/physics/college/wjis45d6fmr2mm0oekxkitx1a7qlzc55px.png)
The change in energy of the transferred charge is given by:
![\Delta U = q \Delta V](https://img.qammunity.org/2020/formulas/physics/high-school/i6295d1ayeusdlw6dq3e8l34qaoleflvjh.png)
where
q is the charge transferred
is the potential difference between the ground and the clouds
Here we have
![q=31 C](https://img.qammunity.org/2020/formulas/physics/college/ms6k4l71q61sfnsqroy0azepe1apwa5fhj.png)
![\Delta V = 0.96\cdot 10^9 V](https://img.qammunity.org/2020/formulas/physics/college/4jve3q09nlxzz62pewwoio896k76ddgwmr.png)
So the change in energy is
![\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^(10) J](https://img.qammunity.org/2020/formulas/physics/college/b901enco34myeqqjok8xunmthegyoz3evp.png)
(b) 7921 m/s
If the energy released is used to accelerate the car from rest, than its final kinetic energy would be
![K=(1)/(2)mv^2](https://img.qammunity.org/2020/formulas/physics/middle-school/c6fs3acuplloc3whu5cpc8ui63cnl7ur39.png)
where
m = 950 kg is the mass of the car
v is the final speed of the car
Here the energy given to the car is
![K=2.98\cdot 10^(10) J](https://img.qammunity.org/2020/formulas/physics/college/d3dw2rvrz00h2uhfo0x6mea4ult3bs8e7g.png)
Therefore by re-arranging the equation, we find the final speed of the car:
![v=\sqrt{(2K)/(m)}=\sqrt{(2(2.98\cdot 10^(10)))/(950)}=7921 m/s](https://img.qammunity.org/2020/formulas/physics/college/nb3xdtd1gmcjf6w0lbfwk9f0m89cqmg3sk.png)