1.99 x 10⁶C
The electrical energy ([tex]E_{E}[/tex]) stored in the battery will give the car some kinetic energy ([tex]E_{K}[/tex]) which will cause the car to move from rest to some other point.
i.e
[tex]E_{E}[/tex] = [tex]E_{K}[/tex] ------------------(i)
But;
[tex]E_{E}[/tex] = [tex]\frac{1}{2}[/tex] x Q x V; -------------------(ii)
Where;
Q = charge on the battery
V = potential difference or voltage of the battery = 12.0V
Also
[tex]E_{K}[/tex] = ([tex]\frac{1}{2}[/tex] x m x v²) - ([tex]\frac{1}{2}[/tex] x m x u²) -----------------(iii)
Where;
m = mass of the car = 956kg
v = final velocity of the car = 158m/s
u = initial velocity of the car = 0 [since the car starts from rest]
Substitute equations (ii) and (iii) into equation (i) as follows;
[tex]\frac{1}{2}[/tex] x Q x V = ([tex]\frac{1}{2}[/tex] x m x v²) - ([tex]\frac{1}{2}[/tex] x m x u²) -----------------(iv)
Substitute all necessary values into equation (iv) as follows;
[tex]\frac{1}{2}[/tex] x Q x 12.0 = ([tex]\frac{1}{2}[/tex] x 956 x 158²) - ([tex]\frac{1}{2}[/tex] x 956 x 0²)
[tex]\frac{1}{2}[/tex] x Q x 12.0 = ([tex]\frac{1}{2}[/tex] x 956 x 158²) - (0)
[tex]\frac{1}{2}[/tex] x Q x 12.0 = ([tex]\frac{1}{2}[/tex] x 956 x 158²)
[tex]\frac{1}{2}[/tex] x Q x 12.0 = 11932792
6Q = 11932792
Solve for Q;
Q = 11932792 / 6
Q = 1988798.67 C
Q = 1.99 x 10⁶C
Therefore, the amount of charge the batteries must have is 1.99 x 10⁶C
