Answer:
The acceleration you can get with that engine in your car is around 70,56
or 7,26
using 1500kg of mass or 3306 pounds
Step-by-step explanation:
Using the equation of the force that is:
![F=m*a](https://img.qammunity.org/2020/formulas/chemistry/middle-school/yfhv56ccnirwl64iwgrk56fhs0yj3nqdf1.png)
So, you notice that you know the force that give the engine, so changing the equation and using a mass of a car in 1500 kg or 3306 pounds
![a=(F)/(m) =(105840 N )/(1500 (kg) )](https://img.qammunity.org/2020/formulas/physics/high-school/ies1dzca6myn18kueymab5owiuddq6p26p.png)
![a=\frac{105840 ((kg*m)/(s^(2) ) )} {1500 kg }](https://img.qammunity.org/2020/formulas/physics/high-school/e1flztwaib4pxdiy890z431f51kfnfn5ya.png)
Note: N or Newton units are:
![(kg * m)/(s^(2) )](https://img.qammunity.org/2020/formulas/physics/high-school/xw1ikwikbfubdr6cfvi8hgiyc0m70gjoof.png)
![a= 70,54 (m)/(s^(2) )](https://img.qammunity.org/2020/formulas/physics/high-school/zu4rxotsziztq0sv2cbie159jn0b249a64.png)
Also in pounds you can compared
![a= (2400 lf )/(3 306 lf)](https://img.qammunity.org/2020/formulas/physics/high-school/va2j2458bc2q1iejhh3m51ihu49mt18hva.png)
Note: lf in force units are:
![(lf*ft)/(s^(2) )](https://img.qammunity.org/2020/formulas/physics/high-school/iqls5ybfxmdvxo0j1yt0op6wofeqj4cqbf.png)
![a=7,26 (ft)/(s^(2) )](https://img.qammunity.org/2020/formulas/physics/high-school/4sfkb3plr2arcfb98btpt7qzg34zu91wng.png)