a) A bus of mass 760 kg requires 120 m to reach certain velocity value Vf. Ignore friction and drag forces and assume the bus engine exerts a constant forward force F. When the bus is towing a 330-kg small car, how long distance needed to reach same Vf? b) If the Vf of the bus is 28 m/s, what is the tension in the tow cable between bus and small car?

Given :  A bus of mass 760 kg requires 120 m to reach certain velocity value Vf.

the bus engine exerts a constant forward force F.

To Find : When the bus is towing a 330-kg small car, how long distance needed to reach same Vf?

Solution:

V²  - U² = 2aS

V = Vf

U = 0

S = 120 m

=> Vf² - 0 = 2a(120)

=> Vf² = 240a

m = 760  kg

Force = F

F = ma

=> F =760 a

=> a = F/760

Vf² = 240F/760

Case 2  :When the bus is towing a 330-kg small car,

m = 760 + 330 = 1090 kg

a = F/1090

Vf²  = 2aS

=> 240F/760  = 2 (F/1090) S

=> S = 120 x 1090 /760

=> S = 172.1   m  

172.1   m   distance needed to reach same Vf

Explanation:

this is a good question! i'll do some research and come back when i have the answer!

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.