The position ModifyingAbove r With right-arrow of a particle moving in an xy plane is given by ModifyingAbove r With right-arrow equals left-parenthesis 3 t cubed minus 2 t right-parenthesis ModifyingAbove i With caret plus left-parenthesis 5 minus 2 t Superscript 4 Baseline right-parenthesis ModifyingAbove j With caret with ModifyingAbove r With right-arrow in meters and t in seconds. In unit-vector notation, calculate
(a)ModifyingAbove r With right-arrow, (b)v Overscript right-arrow EndScripts, and (c)a Overscript right-arrow EndScripts for t = 3 s. (d) What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t = 3 s? Give your answer in the range of (-180o; 180o).
wow - what a lot of words. Just specify stuff with symbols, ok?
r(t) = (3t^3-2t)i + (5-2t^4)j
v(t) = (9t^2-1)i -8t^3 j
a(t) = 18t i - 24t^2 j
(d)
at t=3, dr/dt = v(3) = 26i-216j
so, tanθ = -216/26
To find the answers to the given questions, we need to evaluate the position vector, velocity vector, acceleration vector, and the angle between the positive x-axis and the tangent line at t = 3s.
(a) Position Vector (r):
The position vector, r, is given by:
r = (3t^3 - 2t)i + (5 - 2t^4)j
To find the position vector at t = 3s, substitute t = 3 into the equation:
r = (3(3)^3 - 2(3))i + (5 - 2(3)^4)j
= (81 - 6)i + (5 - 162)j
= 75i - 157j
So, the position vector at t = 3s is r = 75i - 157j meters.
(b) Velocity Vector (v):
The velocity vector, v, is the derivative of the position vector with respect to time (t):
v = dr/dt
Differentiating each component of r separately:
v = (d/dt)(3t^3 - 2t)i + (d/dt)(5 - 2t^4)j
= (9t^2 - 2)i + (-8t^3)j
To find the velocity vector at t = 3s, substitute t = 3 into the equation:
v = (9(3)^2 - 2)i + (-8(3)^3)j
= 77i - 216j
So, the velocity vector at t = 3s is v = 77i - 216j meters per second.
(c) Acceleration Vector (a):
The acceleration vector, a, is the derivative of the velocity vector with respect to time (t):
a = dv/dt
Differentiating each component of v separately:
a = (d/dt)(9t^2 - 2)i + (d/dt)(-8t^3)j
= (18t)i + (-24t^2)j
To find the acceleration vector at t = 3s, substitute t = 3 into the equation:
a = (18(3))i + (-24(3)^2)j
= 54i - 216j
So, the acceleration vector at t = 3s is a = 54i - 216j meters per second squared.
(d) Angle between the positive x-axis and the tangent line:
To find the angle between the positive x-axis and the tangent line at t = 3s, we need to find the slope of the tangent line at that point.
The slope of the tangent line is given by the derivative of the y-component divided by the derivative of the x-component of the position vector:
m = (dy/dt) / (dx/dt)
Differentiating each component of r separately:
dx/dt = 9t^2 - 2
dy/dt = -8t^3
To find the slope at t = 3s, substitute t = 3 into the equations:
dx/dt = 9(3)^2 - 2
= 79
dy/dt = -8(3)^3
= -216
So, the slope of the tangent line at t = 3s is m = (-216)/(79).
To find the angle θ between the positive x-axis and the tangent line, we can use the arctan function:
θ = arctan(m)
Substituting the value of m:
θ = arctan((-216)/(79))
Use a calculator to find the value of θ, which is approximately -68.81 degrees.
Therefore, the angle between the positive x-axis and the tangent line at t = 3s is approximately -68.81 degrees in the range of (-180°, 180°).
To answer this question, we will calculate the position vector (r), velocity vector (v), and acceleration vector (a) using the given equations.
(a) To find the position vector (r), we use ModifyingAbove r With right-arrow = (3t^3 - 2t)i + (5 - 2t^4)j. Plugging in t = 3 s, we get:
ModifyingAbove r With right-arrow = (3(3)^3 - 2(3)i + (5 - 2(3)^4)j
= (81 - 6)i + (5 - 162)j
= 75i - 157j
Therefore, the position vector (r) at t = 3 s is ModifyingAbove r With right-arrow = 75i - 157j meters.
(b) To find the velocity vector (v), we differentiate the position vector (r) with respect to time (t):
v = d/dt (3t^3 - 2t)i + d/dt (5 - 2t^4)j
= (9t^2 - 2)i + (-8t^3)j
Plugging in t = 3 s, we get:
v = (9(3)^2 - 2)i + (-8(3)^3)j
= 73i - 216j
Therefore, the velocity vector (v) at t = 3 s is v = 73i - 216j meters per second.
(c) To find the acceleration vector (a), we differentiate the velocity vector (v) with respect to time (t):
a = d/dt (9t^2 - 2)i + d/dt (-8t^3)j
= (18t)i + (-24t^2)j
Plugging in t = 3 s, we get:
a = (18(3))i + (-24(3)^2)j
= 54i - 216j
Therefore, the acceleration vector (a) at t = 3 s is a = 54i - 216j meters per second squared.
(d) To find the angle between the positive direction of the x-axis and a line tangent to the particle's path at t = 3 s, we can use the angle formula:
θ = arctan(a_y / a_x)
Plugging in the values, we get:
θ = arctan(-216 / 54)
= arctan(-4)
≈ -76.04 degrees
Therefore, the angle between the positive direction of the x-axis and a line tangent to the particle's path at t = 3 s is approximately -76.04 degrees.