It's gonna crash!

We know that $\overrightarrow F=m \overrightarrow a$ .

$m= 1.5$ ton $= 1500$ kg.

$\overrightarrow a = \frac{\triangle \overrightarrow v}{\triangle t}$ .

As the velocity changed from $-10 m/s$ to $2.6 m/s$ ,

the value of $\triangle \overrightarrow v = 12.6$ .

And $\triangle t = 0.2$ as given,

So we get $F= m \frac{\triangle \overrightarrow v}{\triangle t} = \dfrac{1500 \times 12.6}{0.2} = 1500\times 12.6\times 5 = \boxed{94500} N$

Due to the impulse theorem $\triangle p = m \cdot \triangle v = F_m \cdot \triangle t$ Hence $F_m = \frac{m \cdot \triangle v}{\triangle t}=\frac{(1.5\cdot 10^3 kg)(2.6 m/s -(-10 m/s))}{0.2 s} = \boxed{94500 N}$

v = 2.6 m/s ; u = -10 m/s ; t = 0.2 s

Since v = u + at ; a = 63 m/s2

F = ma = (1.5x1000)kg x 63 m/s2 = 94500 N

we can easily compute acceleration further calculate the force exerted by the vehicle ..........