# Homework Help: 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.