Calculating Force and Time for Stopping a Moving Car

[VinnyCee]

Week long Force and Magnitude Problem

I have this problem that I have been working on for about a weel with no progress:

A car weighs 13,000 N and moves at 40 km / h (11.1 m / s) and it stops in 15 m, the force that stops the car is constant. What are the following?:

a) magnitude of the force stopping the car.
b) time required for stop.

I have tried everything I could think of. What equation shold I be using to figure this one?

For b) I have 15 m = (11.1 m/s) (X) s
then x = 1.35 s, right?

For a) I have (1.35 s) / (11.1 m / s) = 8.22 (13,000 N) = 106,089 N force stopping the car, right? (Probably not, but hopefully soomeone can tell me why.)

 

[arildno]

a) Determine the average force from work+energy equation
b) Use that expression for the average force in the impulse+momentum equation to determine t

 

[Diane_]

Part a) Think energy. The car has a certain amount of kinetic energy when it's moving. In order for the car to stop, that energy has to go somewhere, and it has to get there somehow. In this case, there will be a force that acts through a distance to perform work on the car - an energy balance with a little division should get you the answer.

b) I'm not sure where some of your numbers are coming from. You seem to be using the

s = ut + (1/2)at^2 equation, but I don't see enough terms to justify that.

Remember: Under uniform acceleration, you can get the average velocity by averaging the initial and final velocities, both of which you know. You also know that the average velocity is displacement over time - you know the displacement and you need the time. So:

s/t = (u + v)/2

t = 2s/(u + v)

Be sure you keep proper track of your units, as they're all mixed up in this.

 

[VinnyCee]

Thank you Diane,

For b) I know that Avg. Velocity = displacement (15m) divided by time (unkown). But how do I figure the Avg. Velocity?

 

[Diane_]

Again, you can get the average velocity (under uniform acceleration only) by averaging the initial and final velocities. Without a unit change, it'll be 20 kph.

Arildno's solution would work too, of course - it might be worth your while to work it both ways, as you should get the same answer.

 

[arildno]

b) You don't need that at all!
Look at your equation of momentum instead:
Let F be the average force found in a).
Then:
$$F*t=m(v_{final}-v_{initial})$$
or:
$$t=\frac{m(v_{final}-v_{initial})}{F}$$

 

[arildno]

I agree with Diane on this (just so you don't get confused!)

 

[VinnyCee]

Thany you ardilano,

But what is the "work/force" equation that you mentioned for a)?

 

[arildno]

Change of kinetic energy equals work done.
Work equal averge force times distance.

 

[VinnyCee]

I have recalculated and found the avg. velocity to be (11.1 m/s) / (2) = 5.55 m/s, right?

Then I used that in the formula (what is the name of it?) t = (displacement) / (avg. velocity) to get t = (15 m) / (5.55 m/s) and t = 2.7 s, right?

Then I used the formula F = [m*(finalvelocity - initial velocity)] / t
which would be [13000 N * (0-11.1 m/s)] / (2.7 s) and F = 53,444 N, right?

I think that seems kind of high. where did I go wrong?

 

[arildno]

EDIT:
YOU ARE MIXING UP MASS AND WEIGHT!1

 

[VinnyCee]

OIC!

Weight = mass * gravity, right?

Then 13000 N = mass * 9.81 m/s^2 and mass = 1325 kg?

And, F = [(1325 kg)(11.1 m / s^2)] / (2.7 s)
F = 14707.5 N / 2.7 s so then F = 5447 N?

 

[arildno]

In which direction does your force act, relative to your initial velocity?

 

[VinnyCee]

In the opposite direction, right?

so -5447 N?

 

[arildno]

Right!..

 

[VinnyCee]

Thank you all very much! I don't think I could have figured this out without you.

 

[arildno]

Note:
I haven't bothered to check your numbers, only your method of solving.
This looks right, but you might check the actual numbers just to be sure..

 

[VinnyCee]

The units don't match for the last step:

F = [(1325 kg)(-11.1 m / s^2)] / (2.7 s) = (-14707.5 N) / (2.7 s) = -5447 (kg)(m/s)?

 

[arildno]

NO, NO, NO!

Mass unit: kg
Velocity: m/s
Time:s
(kg*(m/s))/s=(kgm)/s^2=N

 

[VinnyCee]

It still does not match (I think):

F = [(1325 kg)(-11.1 m / s)] / (2.7 s) = (-14707.5 (kg)(m/s)) / (2.7 s) = -5447 (kg)(m)?

 

[arildno]

kg*m/s*(1/s)=?+

 

[VinnyCee]

(kg)(m/s)(s) = N

but the end result is only (kg)(m)!

 

[VinnyCee]

Maybe the (kg)(m/s) / (s) = (kg)(m/s^2)?

 

[VinnyCee]

How owuld I find the stopping distance and time if the speed were doubled? What formula would I use?

I wish all of these formulas were in one place for easy reference and application!

 

[arildno]

Maybe the (kg)(m/s) / (s) = (kg)(m/s^2)?

This is correct.

 

[VinnyCee]

I figured it out!

Get the acceleration by divinding the -5447 N force by the mass of the car (1325 kg) to get -4.1 m/s for a of the car.

then find the time by for new stop when the speed is doubled by using a constant acceleration eq: v = (initial velocity) + (acceleration)(time).

Then use the new time to stop and the constant acceleration of -4.1 m/s^2 in this constant acceleration formula: x - x0 = (initialvelocity)(time) + 1/2(acceleration)(time)^2.