jdred23
- 2
- 0
Hello everyone! I've found a physics problem that i don't know the solution of(maybe because of my limited knowledge). The problem is something like this:
Let's say an object travels in a circular path from P to Q and Q to R in which P, Q and Rare not the center of the circle(because P, Q, R are on the circumference of the circle) and we don't know the equation of the circle.
We are given the 2D coordinate of P and Q, the initial velocity at P is 0, the object has the same acceleration all the way from P upto R and the object goes from P to Q in 1 \text{ millisecond} and Q to R in 2 \text{ millisecond}
Is it possible to find the 2D coordinate of R?
The way I tried to solve it:
I managed to find the distance QR by using the following calculations.
We know that:
s_{pq}= (u_{p} * t_{pq}) + (\frac{1}{2}*a_{pq}*t_{pq}^2)
By rearranging we get:
a_{pq} =\frac{\Large{2(s_{pq} - u_{p}*t_{pq})}}{\Large{t_{pq}^2}}\text{...(1)}\\<br /> \text{where } a_{pq} = \text{ acceleration between }P\text{ and }Q, s_{pq} = \text{ the distance between } P \text{ and } Q\text{(found using simple vector math) }, u_{p} = 0\text{( initial velocity is 0 given)} \text{ and }t_{pq} = 1\text{ millisecond(given)}
Now we know a_{pq} Also we know that:
v_{q} = u_{p} + a_{pq} * t_{pq}\text{...(2)} \\<br /> \text{where } v_{q} = \text{ velocity at } Q, u_{p} = 0\text{ (initial velocity)}, a_{pq} = \text{ found above in eq(1) and } t_{pq} = 1\text{ millisecond(given)}
Plugging in a_{pq} and v_q into the equation below we get the distance between Q and R
s_{qr} = v_{q} * t_{qr} + \frac{1}{2} * a_{pq} * t_{qr}^2 \\<br /> \text{where }s_{qr} =\text{ distance between } Q \text{ and } R, v_{q} = \text{ is found above in eq(2) }, a_{pq} = \text{ found above in eq(1) and } t_{qr} = 2\text{ millisecond(given)}
So i know the distance between P and Q and the distance between Q and R.
Now how do I use this information to get the coordinate of R? Any thoughts on this? Is it possible to find the coordinate of R in this way?
Let's say an object travels in a circular path from P to Q and Q to R in which P, Q and Rare not the center of the circle(because P, Q, R are on the circumference of the circle) and we don't know the equation of the circle.
We are given the 2D coordinate of P and Q, the initial velocity at P is 0, the object has the same acceleration all the way from P upto R and the object goes from P to Q in 1 \text{ millisecond} and Q to R in 2 \text{ millisecond}
Is it possible to find the 2D coordinate of R?
The way I tried to solve it:
I managed to find the distance QR by using the following calculations.
We know that:
s_{pq}= (u_{p} * t_{pq}) + (\frac{1}{2}*a_{pq}*t_{pq}^2)
By rearranging we get:
a_{pq} =\frac{\Large{2(s_{pq} - u_{p}*t_{pq})}}{\Large{t_{pq}^2}}\text{...(1)}\\<br /> \text{where } a_{pq} = \text{ acceleration between }P\text{ and }Q, s_{pq} = \text{ the distance between } P \text{ and } Q\text{(found using simple vector math) }, u_{p} = 0\text{( initial velocity is 0 given)} \text{ and }t_{pq} = 1\text{ millisecond(given)}
Now we know a_{pq} Also we know that:
v_{q} = u_{p} + a_{pq} * t_{pq}\text{...(2)} \\<br /> \text{where } v_{q} = \text{ velocity at } Q, u_{p} = 0\text{ (initial velocity)}, a_{pq} = \text{ found above in eq(1) and } t_{pq} = 1\text{ millisecond(given)}
Plugging in a_{pq} and v_q into the equation below we get the distance between Q and R
s_{qr} = v_{q} * t_{qr} + \frac{1}{2} * a_{pq} * t_{qr}^2 \\<br /> \text{where }s_{qr} =\text{ distance between } Q \text{ and } R, v_{q} = \text{ is found above in eq(2) }, a_{pq} = \text{ found above in eq(1) and } t_{qr} = 2\text{ millisecond(given)}
So i know the distance between P and Q and the distance between Q and R.
Now how do I use this information to get the coordinate of R? Any thoughts on this? Is it possible to find the coordinate of R in this way?