Answer:
[tex]t = 8.3 s[/tex]
Explanation:
As we know that
velocity of bike = 7.5 m/s
velocity of car is 10 m/s
deceleration of car is 0.75 m/s^2
part a)
velocity of bike with respect to car is given as
[tex]v_r = 7.5 - 10 = -2.5 m/s[/tex]
acceleration of bike with respect to car is given as
[tex]a_r = 0 - (-0.75) = 0.75 m/s^2[/tex]
now the distance of the bike with respect to car is given as
[tex]d = v_i t + \frac{1}{2}at^2[/tex]
[tex]5 = (-2.5) t + \frac{1}{2}(0.75)t^2[/tex]
[tex]t = 8.3 s[/tex]
Part b)