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