Max Velocity for 1360.7kg Car Climbing 15 Degree Hill

In summary, a 1360.7 kg car with an 80-hp engine that can deliver 59,680 watts to the rear wheels is attempting to climb a 15-degree hill. To find the maximum velocity at which the car can climb, we can use the formula v = Power / mg sin@, where v is the maximum velocity, Power is the engine's power output, m is the mass of the car, g is the gravitational acceleration, and @ is the angle of the hill. Using this formula, the maximum velocity is calculated to be 17.3 m/s.
  • #1
_chris_
8
0
a 1360.7 kg car has an engine which can deliver 59,680 watts to the rear wheels. what is the max. velocity at which the car can climb a 15-degree hill?

i have no idea how to solve this. immediate help would be GREATLY appreciated!
 
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  • #2
oh great. now i read the sticky topic. should have done that before. ha!

well anyway

the work i have done was setting up a triangle with the mass as the hypotenuse...and found the side opposite the 15-degree angle to be 352.17...and from there i am just lost.

i don't see how using the given numbers could help find a maximum velocity. we have a test tomorrow, and this question is on it. our teacher said he would give no advice as to how to solve it. he actually suggested that we find a place much like this one in order to figure out what to do.
 
  • #3
If I assume that the car is climbing up the hill at a constant speed, then

v = s/t where s - distance along the slope.

Energy required for the car to climb up is the gravitational potantial energy mgh = mg s (sin @)

Power = mg s sin@ /t = mg s. sin@ .v/s = mgv sin@

If all the power is delivered for the car, then v max is givne by

v = v max = Power / mg sin@

I don't see any other way to do this.
 
  • #4
Gamma said:
If I assume that the car is climbing up the hill at a constant speed, then

v = s/t where s - distance along the slope.

Energy required for the car to climb up is the gravitational potantial energy mgh = mg s (sin @)

Power = mg s sin@ /t = mg s. sin@ .v/s = mgv sin@

If all the power is delivered for the car, then v max is givne by

v = v max = Power / mg sin@

I don't see any other way to do this.


3451 m/s?

i'm pretty sure i got lost early on there...

if i were to arbitrarily pick some point reasonably far up the slope, would i be able to assume that it had reached its maximum velocity? probably not. bah.
 
  • #5
I am getting, v = 17.3 m/s.

That seems too small.

I assumed that the car initially accelerated quick to a speed vmax and retained this speed during the climb. The problems says, "what is the max. velocity at which the car can climb". It did not say "what is the max speed the car can attain"
 
  • #6
Gamma said:
I am getting, v = 17.3 m/s.

That seems too small.

I assumed that the car initially accelerated quick to a speed vmax and retained this speed during the climb. The problems says, "what is the max. velocity at which the car can climb". It did not say "what is the max speed the car can attain"

how did you get that?? it's much more realistic than my answer. heh. and a lot of answers we get seem to be unrealistic. i dunno...an 80-hp engine is pretty small. but yeah...i have no idea how you came to that answer.
 
  • #7
Plug in the numbers here.

v = Power / mg sin@
 
  • #8
Gamma said:
Plug in the numbers here.

v = Power / mg sin@

that's what i did.

59680 / (1360.7)(9.8)(sin(15)) = ...17.3

i think i screwed up when i put it in the calculator

ha. I'm sorry. that seems good to me. thanks a lot!
 
  • #9
Chris, you can't "set up a triangle with the mass as the hypotenuse ..."

Mass is a scalar, with no direction. Triangle sides are made from vectors,
such as location, velocity, momentum or Force (did you mean weight?)
 

1. What is the maximum velocity that a 1360.7kg car can reach while climbing a 15 degree hill?

The maximum velocity for a car climbing a 15 degree hill depends on several factors such as the power of the car's engine, the aerodynamics of the car, and the condition of the road. However, assuming that the car has a decent engine and is traveling on a paved road, the maximum velocity would be around 55-60 miles per hour.

2. How does the weight of the car affect its maximum velocity while climbing a 15 degree hill?

The weight of the car plays a significant role in determining its maximum velocity while climbing a 15 degree hill. A heavier car will require more power from the engine to climb the hill, resulting in a slower maximum velocity. In this case, a 1360.7kg car would have a lower maximum velocity compared to a lighter car of the same specifications.

3. Is it possible for a car to reach a higher maximum velocity while climbing a 15 degree hill?

Yes, it is possible for a car to reach a higher maximum velocity while climbing a 15 degree hill. This can be achieved by improving the car's engine power, reducing its weight, or using a more aerodynamic design. Additionally, using specialized tires and maintaining a good road condition can also contribute to a higher maximum velocity.

4. How does the angle of the hill affect the maximum velocity of a car?

The angle of the hill has a significant impact on the maximum velocity of a car. As the angle of the hill increases, the force of gravity pulling the car down the hill also increases, making it more difficult for the car to climb. This results in a lower maximum velocity. In this case, a 15 degree hill is considered a moderate incline and would have a moderate effect on the car's maximum velocity.

5. Can a car maintain a constant velocity while climbing a 15 degree hill?

It is technically possible for a car to maintain a constant velocity while climbing a 15 degree hill, but it would require a powerful engine and optimal road conditions. In reality, the car would most likely experience a decrease in velocity as it climbs the hill due to the increased force of gravity. This is why most cars tend to slow down while climbing a hill.

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