How to find the time for a curved path ?

Sdtootle

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))

Lolagoeslala

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

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

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

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

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

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

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

I added comments in blue
So, numerically, what did you get for the speed at C (please show your steps)?

lewando

Please see my edits.

Lolagoeslala

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

Comments in blue
Re-read post #24.

Lolagoeslala

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

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

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

My comments in blue
Are you familiar with free body diagrams? Are you familiar with Newton's first law?

Lolagoeslala

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