Answer :
(a) [tex]2.98\cdot 10^{10} J[/tex]
The change in energy of the transferred charge is given by:
[tex]\Delta U = q \Delta V[/tex]
where
q is the charge transferred
[tex]\Delta V[/tex] is the potential difference between the ground and the clouds
Here we have
[tex]q=31 C[/tex]
[tex]\Delta V = 0.96\cdot 10^9 V[/tex]
So the change in energy is
[tex]\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J[/tex]
(b) 7921 m/s
If the energy released is used to accelerate the car from rest, than its final kinetic energy would be
[tex]K=\frac{1}{2}mv^2[/tex]
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
[tex]K=2.98\cdot 10^{10} J[/tex]
Therefore by re-arranging the equation, we find the final speed of the car:
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s[/tex]