Register to reply

How to find the time for a curved path ?

by Lolagoeslala
Tags: curved, path, time
Share this thread:
Lolagoeslala
#19
Dec22-12, 08:00 PM
P: 217
Quote Quote by Sdtootle View Post
Gravity is the only force acting on the car in the problem you stated. Since there is a ramp, the for is distributed, Part in the direction (g*Cos(theta)) and part in the y direction (g*sin(theta))
So i should use the
y direction (g*sin(theta)) or (g*Cos(theta)) for AD or CD.. well CD would the x component right?
Sdtootle
#20
Dec22-12, 08:21 PM
P: 17
For AD, I would use cos(theta) (the component in the X direction) as it will give you the velocity you need for segment CDo Since there is no acceleration during CD,there are no sin or cos to worry about

Be sure to convince yourself of this as understanding these concepts will make the rest of your semester considerably easier
Lolagoeslala
#21
Dec23-12, 03:30 PM
P: 217
Quote Quote by Sdtootle View Post
For AD, I would use cos(theta) (the component in the X direction) as it will give you the velocity you need for segment CDo Since there is no acceleration during CD,there are no sin or cos to worry about

Be sure to convince yourself of this as understanding these concepts will make the rest of your semester considerably easier
Well when we did this lesson in class. we were looking at the forces... and because it is going down.. we would divide the gravity into gcosθ and gsinθ, so if we are going to a hill ... then i think we would use the angle..... which is between a and c ... which would be 135 deg... and for the gravity component down the slope.. i think it would be (9.8m/s^2)xsin135
Sdtootle
#22
Dec23-12, 11:41 PM
P: 17
If that's the case then it would be because I misunderstood your sketch as the letters weren't entirely clear to me. Regardless, it sounds like you have the right idea. Good luck.
Lolagoeslala
#23
Dec24-12, 02:13 PM
P: 217
Quote Quote by Sdtootle View Post
If that's the case then it would be because I misunderstood your sketch as the letters weren't entirely clear to me. Regardless, it sounds like you have the right idea. Good luck.
So for the straight line path between C and D ... the velocity would be.. the same as A to C right? /.. and the gravity would be 9.8..... so i would make v1 = the constant velocity from A to C ....and find V2 ... and work with that?
lewando
#24
Dec24-12, 02:57 PM
PF Gold
P: 1,053
Quote Quote by Lolagoeslala View Post
So for the straight line path between C and D ... the velocity would be.. the same as A to C right? /.. and the gravity would be 9.8..... so i would make v1 = the constant velocity from A to C ....and find V2 ... and work with that?
The speed at which the car travels path CD is constant.
This speed is equal to the final speed at C, after the car has traversed path AC. The speed along AC is not constant. Initially, at A, it is zero. Do you know how to find the speed at C?
Lolagoeslala
#25
Dec24-12, 03:09 PM
P: 217
Quote Quote by lewando View Post
The speed at which the car travels path CD is constant.
This speed is equal to the final speed at C, after the car has traversed path AC. The speed along AC is not constant. Initially, at A, it is zero. Do you know how to find the speed at C?
yeah i would use the equation V2^2- V1^1 = 2ad

So i would find the V2 = speed from A to C
CD
and then i would use this V2^2 - v1^2 = 2ad
V1^1 = the speed found for AC
V2^2 = need to find

right?
lewando
#26
Dec24-12, 03:18 PM
PF Gold
P: 1,053
I added comments in blue
Quote Quote by Lolagoeslala View Post
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)?
lewando
#27
Dec24-12, 03:36 PM
PF Gold
P: 1,053
Please see my edits.
Lolagoeslala
#28
Dec24-12, 03:41 PM
P: 217
Quote Quote by lewando View Post
I added comments in blue


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
lewando
#29
Dec24-12, 03:54 PM
PF Gold
P: 1,053
Comments in blue
Quote Quote by Lolagoeslala View Post
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 [right]

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.
Lolagoeslala
#30
Dec24-12, 04:16 PM
P: 217
Quote Quote by lewando View Post
Comments in blue

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 [right]...
lewando
#31
Dec24-12, 04:23 PM
PF Gold
P: 1,053
Quote Quote by Lolagoeslala View Post
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 [right]...
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?
Lolagoeslala
#32
Dec24-12, 04:41 PM
P: 217
Quote Quote by lewando View Post
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?
lewando
#33
Dec24-12, 09:14 PM
PF Gold
P: 1,053
My comments in blue
Quote Quote by Lolagoeslala View Post
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?
Lolagoeslala
#34
Dec25-12, 09:24 AM
P: 217
Quote Quote by lewando View Post
My comments in blue

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
Sdtootle
#35
Dec25-12, 09:34 AM
P: 17
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
mfb
#36
Dec25-12, 09:38 AM
Mentor
P: 11,570
Quote Quote by Sdtootle View Post
V = CD * t
Probably just a typo: This should be V=CD/t
Velocity is distance per time.


Register to reply

Related Discussions
An object travels along a curved path. Given acceleration and angle, find velocity. Introductory Physics Homework 9
Find the magnitude of acceleration along a curved path Introductory Physics Homework 0
Differential Equation - Path of Dog Chasing a Rabbit running in a curved path Calculus & Beyond Homework 2
Curved path Introductory Physics Homework 2
Motion Along a Curved Path Introductory Physics Homework 2