Answer:
Approximately [tex]9.7\; \rm m \cdot s^{-2}[/tex].
Explanation:
Assuming that there is no other force on this vehicle, the [tex]16000\; \rm N[/tex] force from the road would be the only force on this vehicle. The net force would then be equal to this [tex]16000\; \rm N\![/tex] force. The size of the net force would be [tex]16000\; \rm N\!\![/tex].
Let [tex]m[/tex] denote the mass of this vehicle and let [tex]\Sigma F[/tex] denote the net force on this vehicle.
By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, [tex]a[/tex], would be:
[tex]\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}[/tex].
For this vehicle, [tex]\Sigma F = 16000\; \rm N[/tex] whereas [tex]m = 1650\; \rm kg[/tex]. The acceleration of this vehicle would be:
[tex]\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}[/tex].
