Car rolling down hill 2-D motion problem

  • Thread starter Maiia
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Homework Statement


Take down as positive. A car is parked near a cliff overlooking the ocean on an incline that makes an angle of 10.5 degrees with the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline and has a velocity of 20m/s when it reaches the edge of the cliff. The cliff is 15.1 m above the ocean. The acceleration of gravity is 9.8m/s2. How far is the car from the base of the cliff when the car hits the ocean? Answer in units of m.

I think I know how to solve the problem, but I'm unsure as to whether the 20m/s they are referring to is the x-component, or the speed, in which case I would need to find the x and y components to solve the problem. I know that in a different scenario, if someone was running then slipped off a cliff that was level, the velocity they were at when they slipped off the cliff would be their x component. But in this case, the car is rolling off an incline... Would it be the same as the other situation?
 

Answers and Replies

  • #2
PhanthomJay
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No, the other situation you refer to, describes a level surface prior to the fall. In this incline surface problem, the 20 m/s is given as a velocity just before leaving the cliff. It has an x and y component.
 
  • #3
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hmm i see. Thanks!
 
  • #4
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i seem to be getting the wrong answer...would you mind looking over my process and telling me if I'm doing something wrong?

I first found the x and y components, then i plugged in the y component and delta y and acceleration into vfy^2= vi^2 + 2ay(delta y) and solved for V final. Then I put Vfinal into the equation Vf= Vi + ayt and solved for t. Then I plugged t and Vx into the equation: delta X= Vx t. I got 2.40363923m
 
  • #5
PhanthomJay
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i seem to be getting the wrong answer...would you mind looking over my process and telling me if I'm doing something wrong?

I first found the x and y components, then i plugged in the y component and delta y and acceleration into vfy^2= vi^2 + 2ay(delta y) and solved for V final. Then I put Vfinal into the equation Vf= Vi + ayt and solved for t. Then I plugged t and Vx into the equation: delta X= Vx t. I got 2.40363923m
that looks ok, you might want to show your numbers.
 
  • #6
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ok so for Vx i got 3.64471051 and for Vy i got 19.66509815. delta y= 15.1 so the first equation after i plugged these numbers in i got Vy= 26.12807083. I plugged Vy and acceleration of 9.8m/s^2 into Vy=V0y+ ayt and got .659487008s as t. Then I plugged t and Vx into delta X= Vxt and fot 2.40363923m. I have an online thing that I'm supposed to submit the answer to, and when I submit it, it shows up as incorrect.
 
  • #7
PhanthomJay
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ok so for Vx i got 3.64471051 and for Vy i got 19.66509815. delta y= 15.1 so the first equation after i plugged these numbers in i got Vy= 26.12807083. I plugged Vy and acceleration of 9.8m/s^2 into Vy=V0y+ ayt and got .659487008s as t. Then I plugged t and Vx into delta X= Vxt and fot 2.40363923m. I have an online thing that I'm supposed to submit the answer to, and when I submit it, it shows up as incorrect.
You have your vx and vy initial mixed up. The incline is 10 degrees with the horizontal. vx is vcos10, vy is vsin10.
 
  • #8
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wow...that was stupid of me -.- Thanks so much for your help :)
 

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