Question

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 2350 kg truck, traveling at a constant speed of 8.6 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 = 2.1 s?
(b) What is the speed of the center of mass of the automobile-truck system then?

Answer 1

Part a)

`d = 14.7 m`

Part b)

`v_f = 8.08 m/s`

Step-by-step explanation:

As we know that initial velocity of the center of mass of the system of truck and car is

`v_(cm) = (m_1v_1 + m_2v_2)/(m_1 + m_2)`

Here we know that

`m_1 = 1050 kg`

`v_1 = 0`

`m_2 = 2350 kg`

`v_2 = 8.6 m/s`

so we have

`v_(cm) = (1050(0) + 2350(8.6))/(1050 + 2350)`

`v_(cm) = 5.94 m/s`

Similarly the acceleration of center of mass is given as

`a_(cm) = (m_1a_1 + m_2a_2)/(m_1 + m_2)`

`a_(cm) = (1050(3.3) + 2350(0))/(1050 + 2350)`

`a_(cm) = 1.02 m/s^2`

Now we know that the center of mass of the system will move a distance

`d = v t + (1)/(2)at^2`

`d = 5.94(2.1) + (1)/(2)(1.02)(2.1)^2`

`d = 14.7 m`

Part b)

final speed of the center of mass is given as

`v_f = v_i + at`

`v_f = 5.94 + (1.02)2.1`

`v_f = 8.08 m/s`