Quick help please!

  • Thread starter _chris_
  • Start date
  • #1
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!
 

Answers and Replies

  • #2
8
0
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
357
11
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
8
0
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
357
11
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
8
0
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
357
11
Plug in the numbers here.

v = Power / mg sin@
 
  • #8
8
0
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
lightgrav
Homework Helper
1,248
30
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?)
 

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