Time travel = 4 s
The car’s final velocity : 18 m/s
Given
initial velocity=vo= 10 m/s
acceleration = a = 2 m/s²
distance=d = 56 m
Required
Time travel and the final velocity
Solution
[tex]\tt d=vo.t+\dfrac{1}{2}at^2\\\\56=10.t+\dfrac{1}{2}.2.t^2\\\\56=10t+t^2\\\\t^2+10t-56=0\\\\(t-4)(t+14)=0\\\\t=4,t=-14\rightarrow take~a~positive~value~t=4~s[/tex]
[tex]\tt v_f=v_i+at\\\\v_f=10+2.4\\\\V_f=18~m/s[/tex]