A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 4.5 m, y = 6.5 m, and has velocity v(naught) =
(6.5 m/s)i-hat + (−2.5 m/s)j-hat (arrow for v(naught) points toward positive x direction). The acceleration
is given by a = (6.5 m/s2)i-hat + (1 m/s2) j-hat (arrow for a points in positive x direction).
1.) What is the x component of velocity after 9 s?
2.) What is the y component of velocity after 9 s?
How do I get this? We went over one example in class and I have 2 formulas to use:
1.) v(subx)=v(naughtx) or (vnaught)x (just realized I took poor notes)
2.) x=v(naughtx)t or x=v(naught)xt
Our class example didn't use i or j hat and didn't ask for x or y after a given time.
Answers given are 65 m/s for x and 6.5 m/s for y; need to figure out how to get them.
To find the x and y components of the velocity after 9 seconds, you can use the equations of motion and the given information.
1.) To find the x component of velocity after 9 seconds:
The equation you should use is:
v(x) = v(naught)x + a(x) * t
Given that the initial velocity v(naught) = (6.5 m/s)i-hat, we can see that the x component of the initial velocity is 6.5 m/s.
Also, the x component of the acceleration a(x) = (6.5 m/s^2)i-hat.
Plugging in the values into the equation:
v(x) = 6.5 m/s + (6.5 m/s^2)i-hat * 9 s = 6.5 m/s + 58.5 m/s = 65 m/s.
Therefore, the x component of the velocity after 9 seconds is 65 m/s.
2.) To find the y component of velocity after 9 seconds:
The equation you should use is:
v(y) = v(naught)y + a(y) * t
Given that the initial velocity v(naught) = (-2.5 m/s)j-hat, we can see that the y component of the initial velocity is -2.5 m/s.
Also, the y component of the acceleration a(y) = (1 m/s^2)j-hat.
Plugging in the values into the equation:
v(y) = -2.5 m/s + (1 m/s^2)j-hat * 9 s = -2.5 m/s + 9 m/s = 6.5 m/s.
Therefore, the y component of the velocity after 9 seconds is 6.5 m/s.
I hope this helps! Let me know if you have any further questions.