Newton's Laws: Simple Constant Force Problem

AI Thread Summary
A car weighing 12,000 N is brought to a stop from an initial speed of 30 km/hr in 4.1 seconds, requiring a constant force to stop. The initial speed converts to 8.33 m/s, and the calculated acceleration is -2.03 m/s², resulting in a stopping force of 2,490 N. The distance covered during braking is calculated using the formula x(t) = x0 + v0*t + 1/2*a*t², but confusion arises regarding the correct time to use. The correct stopping distance is confirmed to be 17.1 m. To solve similar problems, it's essential to convert weight to mass and apply the equations for acceleration and force accurately.
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A car that weighs 12000.0 N is initially moving at a speed of 30.0 km/hr when the brakes are applied and the car is brought to a stop in 4.1 s. Find the magnitude of the force that stops the car, assuming it is constant.

Figured this one out, got: v0 = 8.33 m/s, a = -2.03 m/s^2, F = 2.49 X 10^3 N

What distance does the car move during this time?

This is the one I'm having trouble with.
Using: x(t) = x0 +v0*t + 1/2*a*t^2 and x0 = 0
x(t) = 0 + 8.33(4.6) + 1/2(-2.03)(4.6^2)
x(t) = 16.83 m

Tells me I am wrong. I don't know what to do from here.
 
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"brought to a stop in 4.1 s"

x(4.6) = 0 + 8.33(4.6) + 1/2(-2.03)(4.6^2)

Maybe that's the problem XD
 
Sorry, is the time 4.6 s or 4.1 s?

EDIT: Got beat. LOL.
 
Thanks... :)

17.1 m is the answer.
 
How did you solve the first part? i have the same problem with different numbers and am having trouble with it.
Thanks
 
The formula I used to calculate acceleration was:

a = (v - v0) / t

a is acceleration,
v is final velocity (should be 0 m/s),
v0 is initial velocity, which you must convert to m/s,
t is the time it takes to stop.


You can then use the very important equation

F = ma

to calculate the magnitude of the force that stops the car.

You will need to convert the weight given in Newtons to kilograms. You can do this using the same equation as above, with the acceleration in this equation as the acceleration due to gravity.
 
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