# Homework Help: Problem with acceleration & distance/displacement:

1. Sep 17, 2006

### AznBoi

I have showed my work and gotten an answer but it differs from the actual answer in the back of the book. Thanks for your help!!

A driver in a car traveling at a speed of 60mi/h sees a deer 100m away on the road. Calculate the minimum constant acceleration that is necessary for the car to stop without hitting the deer (assuming that the deer does not move in the meantime).

First I converted 60mi/h into m/s:

(60mi/h) x (1609m/mi) x (1h/3600s) = 26.82 m/s

Then I divided 26.82 by 100m : 100/26.82 = 3.73m/s^2

So it's -3.73m/s^2 because it is slowing down??

Did I do the problem wrong? It seems so short so I think I did.

The acutal answer was -3.6m/s^2 What happened??

2. Sep 17, 2006

### Hootenanny

Staff Emeritus
This step isn't correct, you can check what your doing by using dimensional analysis - check your units;

$$\frac{velocity}{distance} = \frac{m.s^{-1}}{m} = s^{-1} \neq m.s^{-2}$$

You need to use kinematic equations to solve this question. Excellent thread presentation though

3. Sep 17, 2006

### arildno

Take care of your units, you don't get acceleration by dividing a distance with a velocity!
Instead, proceed as follows with x for position, v for velocity and a for acceleration:
$$\frac{dv}{dt}=a\to\frac{dv}{dt}v=av\to\frac{d}{dt}(\frac{1}{2}v^{2})=\frac{d}{dt}(ax)$$
where I've used the fact that "a" is constant.
Thus, integrating between initial time and the time when x=100 and v=0, you get:
$$-\frac{1}{2}(26.82)^{2}=a*100$$

4. Sep 17, 2006

### AznBoi

Oh, so is there a proper kinematic equation for every problem??
I don't know which one to use, am I suppose to use the average acceleration one?

a= (delta)v/(delta)t Cause I'm trying to find the average acceleration right? V would be 26.82m/s , but what is the time?? They don't give it.

Btw, we don't have text books or anything so I have no book source.

What book would you guys recommend me buying to learn Physics B from the start?

I'm a sophomore and I'm in precal.. so I don't know how to use calculus to solve them and I don't understand many things in PHysics!

5. Sep 17, 2006

### arildno

There is only ONE kinematic equation, with varying additional info given for any particular problem.

In your case, that additional info is first and foremost that the acceleration is CONSTANT.

In particular, you'll see that under the assumption of const. acc., you don't NEED the time in order to solve your problem!

Last edited: Sep 17, 2006
6. Sep 17, 2006

### Hootenanny

Staff Emeritus
Last edited by a moderator: Apr 22, 2017
7. Sep 17, 2006

### AznBoi

ok I don't really understand any of the equations that you guy put down but I will look at the sources you gave me. Thanks to you both! =]

Physics is very hard.. I mean not only do you need to use algebra but you need to use your head aswell! a lot too. =[ My brain isn't quite expanding.

8. Sep 17, 2006

### Hootenanny

Staff Emeritus
No problem, if you have any question or need any further help don't hesitate to come back Stick at the physics though

9. Sep 17, 2006

### AznBoi

$$\frac{dv}{dt}=a\to\frac{dv}{dt}v=av\to\frac{d}{dt} (\frac{1}{2}v^{2})=\frac{d}{dt}(ax)$$

I don't get what d- derivative is. I've read about it and it confuses me. Also what are the arrows and how did you get v-velocity to be placed in there.

so how did dv/dt= a -> dv/dt x v?? confusing.. =P

10. Sep 17, 2006

### Hootenanny

Staff Emeritus
http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html#c2

I understand that this stuff may be a bit overwhelming for a pre calc, I would simply recommend learning the formulae without knowing how to derive them before you reach calc.

11. Sep 17, 2006

### AznBoi

ok, I'll see if I can figure it out. Thanks again!

12. Sep 17, 2006

$$\frac{d}{dt}(ax) = \frac{da}{dt}x + a\frac{dx}{dt} = av$$, since $$\frac{dx}{dt}=v$$ and a is constant (so $$\frac{da}{dt}=0$$).