Back with more questions

1. Dec 15, 2004

Kimorto

Thanx for everyone who has helped so far.
(This is not homework, I'm doing problems to study for my final that is in less than 3 hours).
This next question i'm having a hard time with is...

A 2150-kg truck is traveling along a straight,level road at a constant speed of 55.0km/h (15.8m/s) when the driver removes his foot from the accelerator. After 21.0 seconds, the truck's speed is 33.0 km/h (9.2m/s). What is the magnitude of the average net force acting on the truck during the 21.0 second interval?

The answer is 626 N but I have no idea on how to get to it.

2. Dec 15, 2004

marlon

This is one dimensional movement along a straight line...

$$v_f = v_i - a * t$$. where i and f denote initian and final...

just try it and multiply with m to find the force...

regards
marlon

3. Dec 15, 2004

Kimorto

Marlon, that doesn't seem right, I keep getting 675 and the answer says its 626 (multiple choice question, closest answer is 626, and teacher said answer is 626)

4. Dec 15, 2004

marlon

yes, i see your point though i am quite certain of this...

regards
marlon...

perhaps someone else can help...

5. Dec 15, 2004

cyby

F = dp/dt = m(v_2 - v_1)/ t = 2150kg(9.2m/s - 15.8m/s)/21s ~ 675

675 has to be right. Your book/teacher is wrong :P

6. May 19, 2010

poopyphysics

you have your conversions wrong.. 55 km/h is 15.3 m/s not 15.8 m/s. The rest is correct.