Answer:
a ) 21 m b) 7 m/s
Explanation:
For this case, we can consider for our reference system, that point zero (this is, at the rest of the traffic signal) is when X₀ = 0, and direction of movement is from left to right (positive sign)
For the car:
a= acceleration is constant = 5 m/s² m1 , mass = 1100 Kg , V = not constant
This is an uniformly accelerated rectilinear movement, and applicable formulas are:
V = V₀ + at , X = X₀ + V₀t + 1/2at²
For the truck:
V = speed is constant = 7 m/s , mass = 2000 Kg, a = 0
This is an uniform rectilinear movement, and the applicable formula is:
X = X₀ + Vt
a) At t = 3.0 sec, we can use the formula for the track, considering that X₀ = 0 (when it pass the rest of the traffic signal)
X = 0 + (7 m/s)x(3 s) = 21 m
So, at 3 sec, truck will be at 21 m from the rest of traffic light and, as truck has a constant speed; at that exact second; truck and car will be together as a com, so both will be 21 m away from point zero
b) At the exact time (3 sec, or in distance words, 21 m) car and truck will be together as a com, so they will both have the exact speed, hence, speed of car at that point will be 7 m/s
