The equations of motion for a particle are given by: x(t) = 4t and y(t) = 8t^2 + 2. Find the following?

Question

(a) the magnitude and direction of the vector position for t = 2 seconds.

(b) the average vector velocity between 0 and 2 seconds.

(c) the magnitude and direction of the instantaneous vector acceleration for t = 4 seconds.

I'm not sure where to begin because there are two equations instead of just one like I'm used to. How do I start this problem?

? · Accepted Answer

at t = 2 s:
x = 4t = 8 units of length = x-component of P
y = 8t² + 2 = 34 units of length = y-component of P
size of P = √8²+34² = √1220 = 34.9 units of length  ANS (a)
direction of P = arctan y/x = arctan 34/8 = arctan 4.25 = 76.8°  ANS (a)
----
P2 = 34.9 units of length
P0 = 0 units of length
displacement of particle from t=0 to t = 2s is:
d = 34.9 - 0 = 34.9 units of length
time to make this displacement = 2 s
Vavg size = 34.9/2 ≈ 17.5 units of length/s  ANS (b)
Vavg direction = 76.8°  {same direction as P2}  ANS (b)
---
d²x/dt² = 0
d²y/dt² = 16
size of the instantaneous acceleration = 16 units of distance/s²  ANS (c)
direction of the instantaneous acceleration = +Y  ANS (c)

? · Answer

You are on the right track. The instantaneous velocity is the derivative with respect to time of position vector, and accel. is derivative of velocity.

To find the average velocity, take the position at t = 2 and subtract position vector at t = 0, then divide by 2 seconds.

Andrew W · Answer

Is the position vector just r = (4t)i + (8t^2 +2)j ? Because if it is then I feel like an idiot now haha. And is the direction just r(2)?

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The equations of motion for a particle are given by: x(t) = 4t and y(t) = 8t^2 + 2. Find the following?

3 Answers