Answer:
The average acceleration of the driver during the collision is
![-348.4(m)/(s^(2))](https://img.qammunity.org/2020/formulas/physics/high-school/l3o6ydvr01nzlhxlwn9c0xf53edx5zuewh.png)
Step-by-step explanation:
Initial speed of car ,
![u=85(km)/(h)=(85* 5)/(18)(m)/(s)](https://img.qammunity.org/2020/formulas/physics/high-school/l0rmtafu6lvo18z389dtcau3cy7gtwhiqy.png)
=>
![u=23.61(m)/(s)](https://img.qammunity.org/2020/formulas/physics/high-school/l0ybi5o39oqv3s35875bqhzwia0k7x06dj.png)
Finally the car comes to rest .
Therefore final speed of the car ,
![v=0(m)/(s)](https://img.qammunity.org/2020/formulas/physics/high-school/vq0891d2vek4ft05oau7e68ly9nxo3iv0h.png)
Distance traveled while coming to rest , s = 0.80 m
Using equation of motion
![v^(2)=u^(2)+2as](https://img.qammunity.org/2020/formulas/physics/high-school/obiz6wq8wm4hlxtgrpgte68bf8o00xokbq.png)
=>
![0^(2)=23.61^(2)+2* 0.80* a](https://img.qammunity.org/2020/formulas/physics/high-school/10m650ak9ly5g5cgswffuvzvblcls3zl11.png)
=>
![a=-348.4(m)/(s^(2))](https://img.qammunity.org/2020/formulas/physics/high-school/vfd5vj7woehoxuq9sn8hnjtc9v08x2i09z.png)
Thus the average acceleration of the driver during the collision is
![-348.4(m)/(s^(2))](https://img.qammunity.org/2020/formulas/physics/high-school/l3o6ydvr01nzlhxlwn9c0xf53edx5zuewh.png)