Problem with acceleration & distance/displacement:

[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).

My answer is DIFFERENT from the actual answer:

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??

 

[Hootenanny]

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

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

 

[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:
$$-\frac{1}{2}(26.82)^{2}=a*100$$

 

[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 textbooks 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!

 

[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.
Read my previous post.

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

 

[Hootenanny]

If you don't want to use calculus then you are going to have to use kinematic equations (of uniform acceleration); these equations should be included in your textbook. You may want to take a look at https://www.physicsforums.com/showpost.php?p=905663&postcount=2" at Hyperphysics.

 

[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.

 

[Hootenanny]

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.

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

 

[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

 

[Hootenanny]

This page may explain the calculus more clearly;
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.

 

[AznBoi]

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

 

[radou]

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