Answer:
a. 18.56m
b. 8.76m/s
Explanation:
We would be using the equations from Newton's laws to determine the position of the automobile, the automobile speed and the truck's position
Note that :
a = Constant acceleration = 3.3m/s²
t = 2.5s
Automobile position we would use this equation
S = ut + 1/2at²
Where u = 0
S = 0 + 1/2 × 3.3 × (2.5)²
Sa = 10.31m
Automobile speed
v = u + at
Where u = o
v = 0 + (3.3 ×2.5)
v = 8.25m/s
Truck position,
From the question constant speed of the truck = 9.0m/s
S = vt
St = 9.0 × 2.5.
= 22.5m
a. How far is the center of mass of the automobile-truck system from the traffic light at t = 2.5 s?
(MaSa + MtSt)/ Mass of the system
Where Ma = Mass of the automobile = 1050kg
Mt = Mass of the truck = 2200kg
Sa = 10.31m
St = 22.5m
Mass of the system = Mass of automobile + Mass of truck
= 1050 + 2200kg
= 3250kg
= (1050 × 10.31) + (2200 × 22.5)/ 3250
= 60325.5 ÷3250
= 18.56m
b. What is the speed of the center of mass of the automobile-truck system then?
Speed of the center of mass of an automobile-truck system =
= (MaVa + MtVt)/ Mass of the system
Where Ma = Mass of the automobile = 1050kg
Mt = Mass of the truck = 2200kg
Va = Automobile speed= 8.25m/s
Vt = Constant speed of the truck = 9.0m/s
Mass of the system = Mass of automobile + Mass of truck
= 1050 + 2200kg
= 3250kg
= (1050 × 8.25) + (2200 × 9.0)/ 3250
= 28462.5 ÷3250
= 8.76m/s
