A Car Traveling at 29 M/s Runs Out of Gas While Traveling Up a 5.0 âë†ëœ Slope.

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Deceleration on a slope

• Thread starter geejodi
• Start date Feb 16, 2007

Homework Argument

A car traveling at 22.0 m/south runs out of gas while traveling up a 25.0 degree slope. How far up the colina will it coast before starting to curlicue dorsum down?

Homework Equations

Vf^2 = V0^2 + 2ax
10 = V0t + (1/2)at^2 (perhaps?)

The Attempt at a Solution

I am using an online homework system chosen Mastering Physics. I did this trouble and got an answer, which was marked wrong. Here's what I did:

Used Vf^2 = V0^2 + 2ax
where x = vertical position up the gradient.
Vf = 0
V0 = 22sin25
a = -9.8

I solved for x, and got x=4.41 m. This is the vertical altitude, or one leg of the correct triangle across from 25 degrees. I used the law of sines to find the hypotenuse, and got 10.43 meters for the hypotenuse, which would be the distance that the car moved before it stopped.

This was marked wrong. Delight explain why my thinking is wrong, and the correct way to practice the trouble!! Thank you!!

~1000

Answers and Replies

Used Vf^2 = V0^2 + 2ax
where 10 = vertical position upward the slope.
Vf = 0
V0 = 22sin25
a = -9.viii

The acceleration is not 9.8 1000/south^two; that would exist true for a torso in freefall, non something rolling downwards an incline.

Measure out the position, speed, and acceleration parallel to the incline. Then that kinematic formula will make sense.

Lets forget horizontal velocity altogether for a minute. How fast is the auto moving against gravity:

that would be 22*sin(25)=9.30m/due south. From here a few branch points every bit to soln.

using the eqn y'all posted: Vf^ii=Vo^2+2ax
I get 9.3^2/(2*ix.8)="x"=4.41m

And then I agree with the answer, the question is what is asked, past up the colina? then it would be tan 25=4.41/x =9.45m

Information technology seems like y'all are using the wrong acceleration (a = -nine.8). Is the auto in free fall?

EDIT: Md Al trounce me to it.

Lets forget horizontal velocity altogether for a minute. How fast is the car moving confronting gravity:

that would be 22*sin(25)=9.30m/s. From here a few branch points as to soln.

using the eqn you posted: Vf^2=Vo^2+2ax
I get 9.three^2/(2*9.8)="10"=4.41m

So I agree with the answer, the question is what is asked, past up the colina? then it would be tan 25=4.41/ten =9.45m

You lot are bold, like geejodi did, that the acceleration is 9.viii m/southward^2 down. Non and then.

Ok... but how do I observe the acceleration if I don't know the altitude it traveled?

Ok... but how do I observe the acceleration if I don't know the distance it traveled?

By analyzing the forces acting on the machine parallel to the incline, and using Newton'southward 2nd law.

I disagree, there are two ways to arroyo the trouble, rotate the coordinate organization or interruption it down in to a pure Vo(y) which is subject area to -9.8m/south^ii by resolving the vecocity into usual x,y vectors.

And then solving for the vert displacement, become back and get the upwardly the hill displacement. I considered both, this seemed easier to explain.

I disagree, there are two ways to approach the problem, rotate the coordinate system or break information technology downward in to a pure Vo(y) which is subject to -9.8m/s^2 by resolving the vecocity into usual x,y vectors.

And so solving for the vert deportation, go back and get the up the hill displacement. I considered both, this seemed easier to explain.

That is what I did, and it was marked wrong.

I disagree, there are two means to approach the problem, rotate the coordinate organisation or break information technology downwardly in to a pure Vo(y) which is subject to -9.8m/s^2 by resolving the vecocity into usual 10,y vectors.

You are welcome to rotate the coordinate arrangement, merely that won't help. The acceleration of the motorcar is non 9.8 1000/due south^two downwardly. (The incline exerts a force on the car that modifies the dispatch.)

By analyzing the forces acting on the car parallel to the incline, and using Newton's 2d law.

Mass is non given, and we haven't even washed forcefulness withal. I know nearly force caues I did it concluding year but I really don't think this trouble should involve strength.

Mass is not given, and we haven't even done strength yet. I know well-nigh force caues I did it terminal yr simply I really don't call back this problem should involve force.

Perhaps you've covered the fact that the acceleration downwardly an incline (in the absenteeism of friction) is the component of the freefall dispatch along the incline.

Mayhap you've covered the fact that the acceleration down an incline (in the absence of friction) is the component of the freefall acceleration along the incline.

Please oh please, I appreciate your help, but I really wish yous would stop being then cryptic. I breezed through my other physics issues and this one has had me going for an hour. I am wearied and desire to sleep. Delight explicate the concept behind what you are trying to say, clearly? You don't have to requite me the respond, just a concrete style to practise it.... please...

Ok, no thanks to you, I got it. I replaced -9.8 with -9.8sin25. Next fourth dimension don't exist and then cryptic!!! :(

In the absence of friction, I say figure out high loftier could it fly? Answer is simple, upward initial vel less time it takes for gravity to end it. Coud be one caste or 90. same answer. I merely don't go this normal force entering into the eqn? Are yous suggesting that it has no "weight" while traveling up the incline since the World is pushing back? In these cases, I similar to flip to limits, purely up, no normal force, purely horizontal, mg. Lost so lets gets this man in the proper direction, and me too, if we can.

In the absence of friction, I say figure out high high could it fly? Answer is simple, upward initial vel less time it takes for gravity to terminate information technology. Coud be 1 caste or xc. aforementioned answer.

You could but wait at the vertical motion, if you lot like. Take the vertical component of initial velocity combined with the vertical component of the object's dispatch. The dispatch of the object is yard sin(theta) down the incline. The acceleration is not g! (It is not in gratis fall.) The vertical component of acceleration is therefore g [sin(theta)]^two. Apply that and yous will get the right respond. (To me, it's more complicated.)

I merely don't get this normal force entering into the eqn? Are you suggesting that information technology has no "weight" while traveling upwards the incline since the Earth is pushing back?

The acceleration of an object is due to the net forcefulness on it. In this instance two forces act: the normal force and gravity.

Notation to geejodi: I thought I was beingness clear when I said "the acceleration down an incline (in the absence of friction) is the component of the freefall acceleration along the incline". I want you to know where the sin(theta) comes from.

Thanks, I meet your indicate. What I was really advocating now that I call up most my own logic more carefully was setting it up as a potential energy question, i.east separating out the V(y-init) and equating it to the peak H at which it would exist converted to potential energy.

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