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How to find the time for a curved path ?

  • #26
lewando
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I added comments in blue
yeah i would use the equation V2^2- V1^1 = 2ad [V1^2]
So i would find the V2 = speed from A to C [the speed AT C]
CD
and then i would use this V2^2 - v1^2 = 2ad
V1^1 = the speed found for AC [V1^2, speed at A--zero]
V2^2 = need to find the speed at C

right?
So, numerically, what did you get for the speed at C (please show your steps)?
 
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  • #27
lewando
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Please see my edits.
 
  • #28
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So, numerically, what did you get for the speed at C (please show your steps)?
i found the velocity from AC to be :
v2^2 = 2gd
V2 = 44.27188724 m/s

Velocity from C to D or D to C
V2^2 = 2gd + v1^2
v2^2 = 2(9.8)(100) + (44.27188724 m/s)^2
v2 = 62.60990337 m/s
 
  • #29
lewando
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i found the velocity from AC to be :
v2^2 = 2gd [not "g", but g*sin(45), as discussed earlier]
V2 = 44.27188724 m/s


Velocity from C to D or D to C
V2^2 = 2gd + v1^2 [no, don't do this. You have already figured out what the speed is for this interval from the last step.]

Re-read post #24.​
 
  • #30
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Re-read post #24.
But here we are looking at a different distance.... and its gsinθ
I know u guys said to keep the velocity constant which is this 44.27188724 m/s
...​
 
  • #31
lewando
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But here we are looking at a different distance.... and its gsinθ
I know u guys said to keep the velocity constant which is this 44.27188724 m/s
...​


You should avoid doing "...." in your posts. No one knows what that means. Please take the time to complete your thoughts.

Can you please reformulate your last post into one or more questions?​
 
  • #32
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You should avoid doing "...." in your posts. No one knows what that means. Please take the time to complete your thoughts.

Can you please reformulate your last post into one or more questions?
Well, i was thinking since between C and D, the distance is 100 m, not the same as from A to C. And the gravity is not 9.8sinθ, but since its a straight line wouldn't it just be 9.8?
 
  • #33
lewando
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Well, i was thinking since between C and D, the distance is 100 m, not the same as from A to C. [true] And the gravity is not 9.8sinθ, but since its a straight line wouldn't it just be 9.8? [technically speaking, gravity does not change during the course of the car's journey. The acceleration due to gravity affects the motion from A to C, but does not affect the motion from C to D, because of the horizontal nature of the motion]
Are you familiar with free body diagrams? Are you familiar with Newton's first law?
 
  • #34
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Are you familiar with free body diagrams? Are you familiar with Newton's first law?
yes sir i am aware. Oh i see what you are trying to say. If the motion is uniform the gravity does not affect it. But what about the distance difference. Should i just use this equation v2 = at to find the time, not considering the distance change? from CD
 
  • #35
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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
 
  • #36
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V = CD * t
Probably just a typo: This should be V=CD/t
Velocity is distance per time.
 
  • #37
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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
33,888
9,606
Looks correct.
 
  • #39
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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
lewando
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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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