A 1050 kg automobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a constant acceleration of 3.3 m/s2. At the same instant a 2200 kg truck, traveling at a constant speed of 9.0 m/s, overtakes and passes the automobile. (a) How far is the center of mass of the automobile-truck system from the traffic light at t = 2.5 s? (b) What is the speed of the center of mass of the automobile-truck system then?

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

true tan speed = ang vel x radius

increase masses, decrease separation