Explanation:
Given that,
Distance covered, d = 32 m
Time, t = 8.7 m
Final speed of the truck, v = 2.1 m/s
(a) Let a is the acceleration and u is the original speed of the truck.
Using first equation of kinematics as :
[tex]v=u+at[/tex]
[tex]2.1=u+8.7a[/tex]..............(1)
Using second equation of kinematics as :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
[tex]32=u(8.7)+\dfrac{1}{2}a(8.7)^2[/tex]...........(2)
On solving equation (1) and (2) we get :
Original speed, [tex]u=5.25\ m/s[/tex]
(b) Acceleration, [tex]a=-0.36\ m/s^2[/tex]
Hence, this is the required solution.
