How can the speed of a car rolling down a valley be calculated without gas?

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Homework Help Overview

The discussion revolves around calculating the speed of a car that runs out of gas while coasting down a valley. The car starts at a height of 10 meters above the valley floor and needs to reach a gas station located 15 meters above the valley floor on the opposite side.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial conditions of the car's speed and height, and the energy conservation approach used to calculate the final speed. There is a focus on the potential discrepancy in gravitational acceleration values used in calculations.

Discussion Status

Participants are exploring the implications of using different values for gravitational acceleration. There is recognition of the setup being correct, but also acknowledgment of external factors like friction and air drag that may affect the outcome.

Contextual Notes

There is a noted difference in the gravitational acceleration value (9.8 m/s² vs. 10 m/s²) that may influence the calculations. The discussion also hints at the impact of real-world factors such as friction and air resistance on the car's ability to reach the gas station.

port31
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Homework Statement


A 1500kg car traveling at 10m/s suddenly runs out of gas while approaching a valley.
The car is 10m above the valley floor when it starts to coast down the valley.
the gas station is 15meters above the valley floor on the other side.
How fast will the car be going as it coasts to the gas station on the other side?

The Attempt at a Solution


the car is going 10m/s when it coasts and is 10m above the valley floor so it should reach the other side 10m high with a speed of 10m/s. so it will have that much energy to go 5 more meters up to the gas station.
so i set it up like this
[itex]\frac{m{v_i}^2}{2}-mgh=\frac{m{v_f}^2}{2}[/itex]
when I do this i get that the final speed is zero. but my book says it is 1.41m/s
 
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port31 said:

Homework Statement


A 1500kg car traveling at 10m/s suddenly runs out of gas while approaching a valley.
The car is 10m above the valley floor when it starts to coast down the valley.
the gas station is 15meters above the valley floor on the other side.
How fast will the car be going as it coasts to the gas station on the other side?

The Attempt at a Solution


the car is going 10m/s when it coasts and is 10m above the valley floor so it should reach the other side 10m high with a speed of 10m/s. so it will have that much energy to go 5 more meters up to the gas station.
so i set it up like this
[itex]\frac{m{v_i}^2}{2}-mgh=\frac{m{v_f}^2}{2}[/itex]
when I do this i get that the final speed is zero. but my book says it is 1.41m/s

Perhaps the book used a g value of 9.8 rather than 10 ?
 
ya i just realized that. you that seems to be the problem. so my setup is correct.
 
Hi port 31 , welcome to PF!:smile:

Loks like the book used g =9.8m/s^2 and you used g = 10m/s^2. I'd say you are both correct. Considering that with friction and air drag always present, you're not going to make it up the hill anyway.:frown:
 
cragar said:
ya i just realized that. you that seems to be the problem. so my setup is correct.

Your set up was certainly one way of doing it.
 

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