Answer:
10 m
Yes, the car will be able to stop before it reaches the truck.
Explanation:
[tex]u=\text{Initial velocity of the car}=\dfrac{72}{3.6}=20\ \text{m/s}[/tex]
[tex]t=\text{Time}=0.5\ \text{s}[/tex]
[tex]s=\text{Displacement}[/tex]
From the laws of motion we have
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=20\times 0.5+0\\\Rightarrow s=10\ \text{m}[/tex]
The car will travel 10 m in 0.5 seconds
[tex]a=\text{Acceleration}=-2.5\ \text{m/s}^2[/tex]
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -2.5}\\\Rightarrow s=80\ \text{m}[/tex]
Total distance required for the car to stop is 10+80 = 90 m. So, the car will be able to stop before it reaches the truck.
