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

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To calculate the speed of a car rolling down a valley without gas, the initial conditions include a 1500kg car traveling at 10m/s from a height of 10m above the valley floor. The car needs to reach a gas station 15m above the valley floor on the opposite side. The calculations reveal that using a gravitational acceleration of 9.8m/s², rather than 10m/s², leads to a final speed of 1.41m/s, as indicated in the textbook. The discussion highlights the importance of using accurate values for gravitational acceleration and acknowledges that factors like friction and air resistance will affect the car's ability to reach the gas station. Overall, the setup for the energy conservation equation is deemed correct, but the discrepancy in gravitational acceleration is the key issue.
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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
\frac{m{v_i}^2}{2}-mgh=\frac{m{v_f}^2}{2}
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
\frac{m{v_i}^2}{2}-mgh=\frac{m{v_f}^2}{2}
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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