A particle's trajectory is described by x =(12t3−2t2)m and y =(12t2−2t)m, where t is in s.
What is the particle's speed at t=0s ?
What is the particle's speed at t=5.0s ?
Express your answer using two significant figures.
What is the particle's direction of motion, measured as an angle from the x-axis, at t=0 s ?
Express your answer using two significant figures.
What is the particle's direction of motion, measured as an angle from the x-axis, at t=5.0s ?
Express your answer using two significant figures.
A. v=2m/s
C. -90
Henry’s post is incorrect. Instinctively, it might be easier to just plug in the values for t, but in order to find speed from a position equation, you must take the derivative.
V(x) = x ‘(t)
V(y) = y ‘(t)
Then take the magnitude of both components. It may help to draw this on a graph, perhaps a position-time graph and / or velocity-time graph.
B. 17.75
Because
sqrt([3/ 2 t^2 - 4t]^2 + [t - 2]^2).
x =(1/2t^3−2t^2)m and y =(1/2t^2−2t)m
To find the particle's speed at a given time, we need to find the magnitude of its velocity vector. The velocity vector is the derivative of the position vector with respect to time.
Given that the particle's position vector is x = (12t^3 - 2t^2) m and y = (12t^2 - 2t) m, we can find the velocity vector by taking the derivative of each component with respect to time.
For the x-component:
dx/dt = d/dt (12t^3 - 2t^2)
= 36t^2 - 4t
For the y-component:
dy/dt = d/dt (12t^2 - 2t)
= 24t - 2
The velocity vector is therefore v = (36t^2 - 4t) i + (24t - 2) j, where i and j are the unit vectors in the x and y directions, respectively.
Now, we can calculate the speed of the particle at a given time by finding the magnitude of the velocity vector.
At t = 0s:
v = (36(0)^2 - 4(0)) i + (24(0) - 2) j
= -2 j
The speed at t = 0s is the magnitude of the velocity vector, which is |v| = | -2 j | = 2 m/s.
At t = 5.0s:
v = (36(5.0)^2 - 4(5.0)) i + (24(5.0) - 2) j
= 830 i + 118 j
The speed at t = 5.0s is |v| = |830 i + 118 j| = √(830^2 + 118^2) = 839 m/s (rounded to two significant figures).
To find the direction of motion at a given time, we need to find the angle between the velocity vector and the x-axis. We can use trigonometry to calculate this angle.
At t = 0s:
The direction of motion is given by the angle θ = tan^(-1)(v_y / v_x), where v_x and v_y are the x and y components of the velocity vector, respectively.
θ = tan^(-1)((24(0) - 2) / (36(0)^2 - 4(0)))
= tan^(-1)(0 / 0)
= undefined
At t = 5.0s:
θ = tan^(-1)((24(5.0) - 2) / (36(5.0)^2 - 4(5.0)))
= tan^(-1)(118 / 830)
≈ 7.99° (rounded to two significant figures).
Therefore, at t = 5.0s, the particle's direction of motion, measured as an angle from the x-axis, is approximately 8.0°.
Plug it in
A. Speed = 0.
B. X = 12*5^3 - 2*5^2 =
Y = 12*5^2 - 2*5 =
Speed = sqrt(X^2 + Y^2) =
C.
D. Tan A = Y/X.
A = ?