Determine the acceleration of the truck in terms of a.

Let $m_{car}$ be the mass of the car and $m_{truck}$ be the mass of the truck.
Let $a_{car}$ be the acceleration of the car and $a_{truck}$ be the acceleration of the truck.
Assume $a$ refers to the acceleration of the car, so $a_{car} = a$.

According to Newton's Third Law, the magnitude of the force exerted by the car on the truck ($F_{ct}$) is equal to the magnitude of the force exerted by the truck on the car ($F_{tc}$). Let this magnitude be $F$.

$$F_{ct} = F_{tc} = F$$

From Newton's Second Law ($F = ma$):
For the car: $F = m_{car} a_{car}$
For the truck: $F = m_{truck} a_{truck}$
Equating the forces:

$$m_{truck} a_{truck} = m_{car} a_{car}$$

Substitute $a_{car} = a$:

$$m_{truck} a_{truck} = m_{car} a$$

Solve for $a_{truck}$:

$$a_{truck} = \frac{m_{car}}{m_{truck}} a$$

The acceleration that the truck experiences on impact is $\boxed{\text{}\frac{m_{car}}{m_{truck}} a\text{}}$

Reason: According to Newton's Third Law, the force exerted by the car on the truck is equal in magnitude to the force exerted by the truck on the car. Since the truck typically has a much larger mass than the car ($m_{truck} > m_{car}$), for the same force, the truck's acceleration will be smaller than the car's acceleration.

The $\boxed{\text{car driver}\text{}}$ is likely to be more severely injured on impact.

Explanation:
During the collision, the less massive car experiences a greater change in velocity ($\Delta v$) compared to the more massive truck, for the same impulse. Since the time of impact ($\Delta t$) is the same for both vehicles, the car undergoes a significantly larger deceleration ($a = \frac{\Delta v}{\Delta t}$). This greater deceleration means a larger force is exerted on the car driver by the seatbelt, leading to more severe injuries.