# Decelleration problem

1. Mar 2, 2005

### Power24

Driver travelling at 25 m/s when she spots a sign that reads "Bridge Out Ahead". It takes her 1.0s to react and begin braking. Car slows down at a rate of 3.0m/s^2. She stops 5.0m short of thewashed out bridge.

A) How much time was required to stop once the brakes were applied?
B) How far from the bridge was she when she first noticed the sign?

I think I can get B as soon as someone helps me with A. I have no idea what equation to use first off because I have tried what I believed was the right one without any success.

Vi= 25 m/s
Vf= 0m/s
acelleration rate= 3.0 m/s^2
Time = ?

Logically the answer will be approximately 2.6, just looking for the right formula please.

Last edited: Mar 2, 2005
2. Mar 2, 2005

### bjr_jyd15

remember the basic accel. equ?

a=(Vf-Vi)/t

You know accel., and both initial and final speeds, so plug it in to find time.

3. Mar 2, 2005

### Jameson

$$d = v_{i}t + \frac{1}{2}at^2$$

You know d, vi, and a. Solve for "t". And then add 1 for the second it took her to react.

EDIT: Use the formula for part 2. You don't know the distance for part A. My bad.

Last edited: Mar 2, 2005
4. Mar 2, 2005

### Power24

I guess my problem lies within the acceleration, the ^2 has me thrown off on just exactly how to figure it out. I'm also having trouble re-writing these formulas, In the formulas I've been given in the book they refer to the average acceleration. It just seems to me that 3.0m/s^2 is not an average acceleration rate, it's just the rate overall.

Time=25-0/3.0^2
then I get
Time=0.083

Last edited: Mar 2, 2005
5. Mar 2, 2005

### bjr_jyd15

well think of acceleration as m/s/s. you lose 3 m/s of speed every second.

also remember a=-3.0 m/s/s because it is decelerating.

6. Mar 2, 2005

### bjr_jyd15

working equation is t=(Vf-Vi)/a

plug in t=(0-25 m/s)/(-3.0 m/s/s) = 8.3 s

7. Mar 2, 2005

### Jameson

The rate overall? I would think in this type of question they would give you the average acceleration.

And sorry, I gave you the wrong equation to work with. You don't get the distance it took her to stop.

Use the one bjr_jyd15 gave you.

$$a = \frac{(v_{f}-v_{i})}{2}*t$$

Is there a way to check the answer?

8. Mar 2, 2005

### Power24

There is no way to check the answer. I'll try working with the information I've been givien and come back in a few minutes.

Well thanks for the help, it appears that 8.3seconds is the right answer to me. I've done it on paper to come to that answer. It just doesn't make sense to me logically.

Last edited: Mar 2, 2005
9. Mar 2, 2005

### bjr_jyd15

Power, tell me why it doesn't make sense? FYI, 25 m/s = 56 mph, so it seems reasonable. If you have trouble with part B, let me know. Remember you are solving for distance. (Think of an equation in which distance is already solved for)

10. Mar 2, 2005

### Jameson

I got the same thing, $8.33333$ or $\frac{25}{3}$

You forgot to add 1s for the time it took her to react however.

Look at it this way. She was going 25 m/s and slowed down at a rate of -3.0m/s^2.
You found the time it took to be 8.333. If you multiply the average acceleration (-3) by the amount of time she slowed down (8.333), you get 25, which is how fast she was going originally.

How are you going to about the second part?

11. Mar 2, 2005

### bjr_jyd15

So time is 8.3 seconds for PART A.

PART B asks when she first noticed the sign. So now add the 1 s. t=9.3 s.

Remember the formula Jameson gave you above (like the third post). Try using that!

12. Mar 2, 2005

### Power24

The reason it didn't make sense to me is because I was looking at it like this,

1second = 3m - 25m/s
2seconds = 9m - 25m/s
3seconds = 27m - 25m/s

and yes I was going to use the formula provided in the third post for part b.

Last edited: Mar 2, 2005
13. Mar 2, 2005

### Jameson

Those are different units.... anyway, do you see how it works logically? And again, how are you going to solve the second part? :yuck:

14. Mar 2, 2005

### bjr_jyd15

Can you explain what you mean by that, Power?

Is it supposed to be 3m/s?

15. Mar 2, 2005

### Power24

What I meant by that is in the question it gives 3.0m/s^2, I interpreted 3.0m/s^2 to mean that every second the value of 3 is powered to the 2nd.

I can see now that ^2 does not apply to numerical value but the m/s.

here is how I've done part b

d= (25)(9.3)+1/2(3)(9.3)^2
d=232+127.9
d=359.9

I have a feeling thats the wrong answer though...
But I have corrected myself, the acceleration is -3.0 not positive, even though you have mentioned this earlier.

That now gives me an answer of d= 104.1

Last edited: Mar 2, 2005
16. Mar 2, 2005

### Jameson

The way you did the question implies she braked for 9.333 seconds. This is a tricky question. The total distance she traveled was.

$$d = (25)(\frac{25}{3})+\frac{1}{2}(-3)(8.3)^2$$

This gives you the distance it took her to stop.... she went for 25 m/s for 1 second before she stopped.

So add 25(1)m to that distance + 5 m to the bridge.

17. Mar 2, 2005

### Power24

hmmm, touchet. I now however grasp how to do the question. Much thanks for all the help.

Part B = 134.165

Last edited: Mar 2, 2005