Answer:
7.15 m/s
Step-by-step explanation:
We use a frame of reference in which the origin is at the point where the trucck passed the car and that moment is t=0. The X axis of the frame of reference is in the direction the vehicles move.
The truck moves at constant speed, we can use the equation for position under constant speed:
Xt = X0 + v*t
The car is accelerating with constant acceleration, we can use this equation
Xc = X0 + V0*t + 1/2*a*t^2
We know that both vehicles will meet again at x = 578
Replacing this in the equation of the truck:
578 = 24 * t
We get the time when the car passes the truck
t = 578 / 24 = 24.08 s
Before replacing the values on the car equation, we rearrange it:
Xc = X0 + V0*t + 1/2*a*t^2
V0*t = Xc - 1/2*a*t^2
V0 = (Xc - 1/2*a*t^2)/t
Now we replace
V0 = (578 - 1/2*1.4*24.08^2) / 24.08 = 7.15 m/s