Explanation:
Given that,
Initial speed of the car, u = 88 km/h = 24.44 m/s
Reaction time, t = 2 s
Distance covered during this time, [tex]d=24.44\times 2=48.88\ m[/tex]
(a) Acceleration, [tex]a=-4\ m/s^2[/tex]
We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{-(24.44)^2}{2\times -4}[/tex]
s = 74.66 meters
s = 74.66 + 48.88 = 123.54 meters
(b) Acceleration, [tex]a=-8\ m/s^2[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{-(24.44)^2}{2\times -8}[/tex]
s = 37.33 meters
s = 37.33 + 48.88 = 86.21 meters
Hence, this is the required solution.
