Question

A particle's position r as a function of time t is given by r(t) = at^3 i + (bt-ct^4) j where a = 5.00 m/s^3, b = 3.00 m/s, and c = 6.00 m/s^4. At t = 2.47 s find: the x component of position, the y component of position, the magnitude of the vector that points from the origin to the particle, the direction of the vector that points from the origin to the particle, the x component of velocity, the y component of velocity, the magnitude of the velocity vector, the direction of the velocity vector, the x component of the acceleration, the y component of the acceleration, and the magnitude of the acceleration.

Answer 1

Wow!! you posted the whole problem and asked for all?
first off, replace a, b,c I get the function of \(r(t)= 5t^3 \hat {i} +(3t - 6t^4)\hat{j}\) . So, at t = 2.47s,
a) the x component of position is 5*(2.47)^3 = 75.3 m
b) the y component of position is 3*2.47 + 6*(2.47)^4 = 230.74m
c) the magnitude of the vector from the origin is \(\sqrt{(75.3)^2 + (230.74)^2}= 242.71\)m
to the velocity, you have to take derivative of r(t). It is \(\vec v = 15t^2 \hat {i} + (3-24t^3)\hat {j}\)
At t = 2.47s
a) the x component of velocity is 15* (2.47)^2 = 91.5 (m/s)
b) the y component of velocity is 3 - 24(2.47)^3 = -358.7 m/s
c) the magnitude of vector velocity is \(|\vec v| = \sqrt {(91.5)^2 + (-358.7)^2} = 370.2m\)

Answer 2

Looks good, just missing direction for the position and velocity vector and the magnitude of the y-component of acceleration and its total magnitude -- but you have everything needed to get these calculated.

Answer 3

The direction for velocity is calculated as \(\large {\tan\theta=\dfrac{v_y}{ v_x}} \) and similarly for the position vector.

Answer 4

$$

\theta_{velocity}=\arctan \frac{v_y}{v_x}=\arctan \frac{-358.7}{ 91.5}=-75.7^\circ

$$

Answer 5

yes,75.7 degrees below horizontal