- #1

niyati

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**a**= 3.00j m/s^2 and an initial velocity of

**v[initial]**= 500i m/s. Find (a) the vector position and velocity at any time

*t*and (b) the coordinates and speed of the particle at

*t*= 2.00s.

(a) I looked at the examples given in the chapter, and to find the velocity at any time seemed to be the final velocity, which was found through the equation of

**v[f]**=

**v**

*+*

ax = 0 ay = 3.00 vx = 500 vy = 0

Working each component separately, I got 500i + (3.00t)j.

For the position, I, again, separately worked each component, using the equation

**a***t*. That means that the final velocity is equal to it's components added, and the same goes for the initial velocity and acceleration. So:ax = 0 ay = 3.00 vx = 500 vy = 0

Working each component separately, I got 500i + (3.00t)j.

For the position, I, again, separately worked each component, using the equation

**r[f]**+**r***+***v**

The initial position is (0,0), so my answer is:

(b) To find the coordinates of position, I took my last equation from above, and broke it down into it's components, where x[f] = x*t*+ .5**a***t^2*.The initial position is (0,0), so my answer is:

**r[f]**= (500t)i + (1.5(t^2))j(b) To find the coordinates of position, I took my last equation from above, and broke it down into it's components, where x[f] = x

*+ vx**+ .5(ax)(t^2) and y[f] = y**+ vy**+ .5(ay)(t^2), all while plugging in 2 for time. x = 1000 and y = 3.*

For speed, my previous findings of 500i + (3.00t)j, I got 500i + 6.00j when plugging in 2. That would be the velocity, and taking the absolute value would equal speed. But, I don't think speed is in it's components, and so I drew the hypotenuse of the triangle the vectors formed, and got 500.036 m/s.

...for some reason, I'm not sure if that is right. I kind of guess on taking the hypotenuse, but I don't know if I should do that, if I've already plugged into an equation specifically for velocity. What I'm wondering seems stupid, but, well, X/

Thank you in advance.For speed, my previous findings of 500i + (3.00t)j, I got 500i + 6.00j when plugging in 2. That would be the velocity, and taking the absolute value would equal speed. But, I don't think speed is in it's components, and so I drew the hypotenuse of the triangle the vectors formed, and got 500.036 m/s.

...for some reason, I'm not sure if that is right. I kind of guess on taking the hypotenuse, but I don't know if I should do that, if I've already plugged into an equation specifically for velocity. What I'm wondering seems stupid, but, well, X/

Thank you in advance.