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A 950 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 2.6 m/s2. At the same instant a 1950 kg truck, traveling at a constant speed of 6.8 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 = 3.0 s?
b. What is the speed of the com then?

Respuesta :

Khoso123

Answer:

(a) [tex]x_{sys}=17.55m[/tex]

(b) [tex]v_{sys}=6.833m/s[/tex]

Explanation:

Given data

950 kg automobile starts from rest with constant acceleration  2.6m/s²

1950 kg truck, traveling at a constant speed of 6.8 m/s

Automobile position

[tex]x_{a}=1/2at^{2}\\x_{a}=1/2(2.6m/s^{2})(3)^{2}\\x_{a}=11.7m[/tex]

Velocity

[tex]v_{a}=at\\v_{a}=(2.3m/s^{2})(3.0s)\\v_{a}=6.9m/s[/tex]

Truck Position

[tex]x_{t}=vt\\x_{t}=(6.8m/s)(3.0s)\\x_{t}=20.4m[/tex]

Velocity of truck remain constant vt=6.8m/s

For Part (a)

The center of mass for Automobile Truck system:

[tex]x_{sys}=\frac{m_{a}x_{a}+m_{t}x_{t}}{m_{sys}}\\x_{sys}=\frac{(950kg*11.7m)+(1950kg*20.4m)}{950kg+1950kg}\\ x_{sys}=17.55m[/tex]

For Part (b)

The velocity of center of mass for Automobile Truck system:

[tex]v_{sys}=\frac{m_{a}v_{a}+m_{t}v_{t}}{m_{sys}}\\v_{sys}=\frac{950*6.9+1950*6.8}{950+1950}\\ v_{sys}=6.833m/s[/tex]

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