How to find the time for a curved path ?

  • Thread starter Lolagoeslala
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In summary, the car travels a distance of 9.8 meters per second and takes 52.6484433 milliseconds to complete the journey.
  • #36
Sdtootle said:
V = CD * t
Probably just a typo: This should be V=CD/t
Velocity is distance per time.
 
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  • #37
Sdtootle said:
If you are able to find the final velocity at the end of the ramp (I'll call V), then the time from C to D can be found with:
V = CD * t
Where CD is the distance between CD. When on a friction less, horizontal plain; the force due to gravity is equal and opposite to the normal force and no acceleration takes place. Since the plain lacks friction, there is no deceleration. Therefore, velocity stays constant during CD

So like this :

V= 44. 277188724 m/s (this is the velocity from AC)
D = 100 m (CtoD distance)
t = (time)
t= 100 m / 44. 277188724 m/s
t = 2.258769757 s
 
  • #38
Looks correct.
 
  • #39
Lolagoeslala said:
So like this :

V= 44. 277188724 m/s (this is the velocity from AC)
D = 100 m (CtoD distance)
t = (time)
t= 100 m / 44. 277188724 m/s
t = 2.258769757 s

Thanks mfb! I can be so slow in the mornings.

Lola: don't forget to account for the time going down the ramp.

Total time should be closer to 3s, but still slightly less then 3s.
 
  • #40
I agree with Sdtootle.

The key idea here is to note that the time it takes to travel from A to D is the same time that the car's "shadow" moves from B to D. t(AD) = t(BD).

Then 100/v = 100(sq root 2). Solve for v. Then t(total journey) = D/v .
 
  • #41
tomstringer said:
I agree with Sdtootle.

The key idea here is to note that the time it takes to travel from A to D is the same time that the car's "shadow" moves from B to D. t(AD) = t(BD).
This may be true, but how does it help?
Then 100/v = 100(sq root 2). Solve for v. Then t(total journey) = D/v .
Not following what you are trying to say. Solving for v this way, you get v = 1/√2 m/s. I thought we established that the speed at the bottom of the hill is 44.27 m/s.
 

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