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:

well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.