Average Net Force

1. Sep 26, 2006

brncsfns5621

An 810-kg car accelerates from rest to 27m/s in a distance of 120m. What is the magnitude of the average net force acting on the car?

120m / 27m/s= 4.4 s

F=ma
=810kg * 4.4s
= 3600 N

This is not one of my choices. What did I do wrong?

2. Sep 26, 2006

You can not apply t = s / v, because the car is accelerating. The velocity of the car is given with 27 = a * t. The second equation is the displacement of the car, 120 = 1/2*a*t^2. Now, plug in the first equation into the second one, and get the time and acceleration. Then you can calculate the average force acting on the car.

3. Sep 26, 2006

brncsfns5621

So, using your equations I get:

120m = 1/2 (27m/s)^2
120m = 1/2 (729m^2/s^2)
120m = 364m^2/s^2
rearrange--> s^2 = 364m^2/120m
s^2 = 3.03m

This isn't right....

4. Sep 26, 2006

brncsfns5621

Anyone have any info on this?

5. Sep 26, 2006

You have two equations:

(1)... 27 = a*t
(2)... 120 = 1/2*a*t^2

From (1) you obtain a = 27/t. Putting that into (2) gives t = 8.89 seconds, which implies (from (1) ) a = 3.04 m/s^2. Hence, F = m*a implies F = 810 * 3.04 = 2462.4 [N]. Do not mix units with variables. You do not have tu put units (such as m/s, for example) into your equations. In other words, you wrote
which reads: 'Seconds squared equals 3.03 meters.' Does not make any sense, right?

Last edited: Sep 26, 2006
6. Sep 26, 2006

brncsfns5621

WOW!!! I brain farted that one. For some reason I was thinking to replace the "a*t" with 27. Don't know why... Thank you for your help, I really appreciate it.

7. Jan 13, 2008

rainingurl

solve the problem

a = ?
a = v/t
solve t using d=.5(v-V)t
t= 8.8sec

sub in the equation
a = (27m/s)/8.8sec
a = 3.06m/s^2

F= mass*acceleration
810kg*3.06m/s^2 = 2485N = 2500N

8. Dec 18, 2009

dkzdrood

KE = 1/2 * m v^2
KE = (1/2) (810) (27^2)
W= deltaKE
W = (1/2) (810) (27^2)
W = force * distance
(1/2)(810)(27^2) = 120*F
F= [(1/2)(810)(27^2)] / (120)
F= 2460.375 N