Answer
$a_{max}=16.13m/s^2$
Work Step by Step
The car can speed up without its tires slipping if the net force propelling the car to move does not exceed the down force, which is $4060N$ here.
$$\sum F\le4060N$$
There are 3 forces acting on the car when it starts to speed up:
- The propelling force from the engine $P$
- The static frictional force stopping the car from speeding up $f_s$
- Air resistance force $F_a=1190N$, also working against the car's motion
$$\sum F=P-f_s-F_a\le4060N$$
$f_s$ can be calculated: $f_s=\mu_sF_N=\mu_s\times m_{car}g=0.87\times690\times9.8=5883N$
So, $$P\le4060+5883+1190$$ $$P\le11133N$$
From Newton's 2nd Law, $P=m_{car}a$ with $a$ being the acceleration of the car
$$m_{car}a\le11133N$$ $$a\le16.13m/s^2$$
Therefore, $a_{max}=16.13m/s^2$
