A car is having weight W is moving in the direction as shown in the figure. The centre of gravity (CG) of the car is located at height h from the ground, midway between the front and rear wheels. The distance between the front and rear wheels, is l. The acceleration of the car is a, and acceleration due to gravity is g. The reactions on the front wheels (R_f ) and rear wheels (R_r ) are given by

To find the normal reactions on the front wheels ($R_f$) and rear wheels ($R_r$), we can analyze the forces acting on the car.

Let the mass of the car be
$m=\frac{W}{g}$
where $W$ is the weight of the car and $g$ is the acceleration due to gravity.

The car is accelerating to the right with an acceleration $a$. In the frame of reference of the accelerating car, we can apply a D'Alembert's pseudo-force (inertia force) acting horizontally in the backward direction (to the left) at the Centre of Gravity (CG).
The magnitude of this pseudo-force is:
$F_i=ma=\frac{W}{g}a$

The Centre of Gravity (CG) is located at a height $h$ from the ground, midway between the wheels. Since the wheelbase distance is $l$, the horizontal distance from the CG to both the front and rear wheel contact points is:
$d=\frac{l}{2}$

Taking moments about the contact point of the rear wheels (to eliminate $R_r$):
$\sum M_{\text{rear}}=0$
$(R_f\times l)-(W\times \frac{l}{2})+(F_i\times h)=0$
Substituting $F_i=\frac{W}{g}a$:
$R_f\times l=\frac{Wl}{2}-\frac{W}{g}ah$
Dividing both sides by $l$:
$R_f=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{l}\right)a$

Next, taking moments about the contact point of the front wheels (to eliminate $R_f$):
$\sum M_{\text{front}}=0$
$-(R_r\times l)+(W\times \frac{l}{2})+(F_i\times h)=0$
$R_r\times l=\frac{Wl}{2}+\frac{W}{g}ah$
Dividing both sides by $l$:
$R_r=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{l}\right)a$

Thus, the reactions on the front wheels ($R_f$) and rear wheels ($R_r$) are:
$R_f=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{l}\right)a$
and
$R_r=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{l}\right)a$