The position of a particle moving along an x axis is given by x = 15t2 - 2.0t3, where x is in meters and t is in seconds.
(a) Determine the position, velocity, and acceleration of the particle at t = 3.0 s.
x = m
v = m/s
a = m/s2
(b) What is the maximum positive coordinate reached by the particle?
m
At what time is it reached?
s
(c) What is the maximum positive velocity reached by the particle?
m/s
At what time is it reached?
s
(d) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)?
m/s2
(e) Determine the average velocity of the particle between t = 0 and t = 3 s.
m/s
not good
(a) To determine the position, velocity, and acceleration of the particle at t = 3.0 s, we substitute t = 3.0 s into the given equation.
x = 15t^2 - 2.0t^3
x = 15(3.0)^2 - 2.0(3.0)^3
x = 135 - 2.0(27)
x = 135 - 54
x = 81 meters
So, the position of the particle at t = 3.0 s is 81 meters.
To find the velocity, we take the derivative of the position function with respect to time (t):
v = dx/dt = d(15t^2 - 2.0t^3)/dt
v = 30t - 6.0t^2
Substituting t = 3.0 s:
v = 30(3.0) - 6.0(3.0)^2
v = 90 - 54
v = 36 m/s
So, the velocity of the particle at t = 3.0 s is 36 m/s.
To find the acceleration, we take the derivative of the velocity function with respect to time (t):
a = dv/dt = d(30t - 6.0t^2)/dt
a = 30 - 12.0t
Substituting t = 3.0 s:
a = 30 - 12.0(3.0)
a = 30 - 36
a = -6 m/s^2
So, the acceleration of the particle at t = 3.0 s is -6 m/s^2.
Therefore, at t = 3.0 s:
x = 81 m
v = 36 m/s
a = -6 m/s^2
(b) To find the maximum positive coordinate reached by the particle, we need to find the maximum value of x. To do this, we can find the derivative of x with respect to t and set it equal to zero:
v = dx/dt = d(15t^2 - 2.0t^3)/dt
v = 30t - 6.0t^2
To find the maximum value of x, we set v = 0 and solve for t:
30t - 6.0t^2 = 0
t(30 - 6.0t) = 0
This equation has two solutions:
t = 0 (corresponds to the initial position)
t = 5 s
To determine which of these values gives the maximum positive coordinate, we substitute them into the position equation:
x = 15t^2 - 2.0t^3
Substituting t = 0:
x = 15(0)^2 - 2.0(0)^3
x = 0
Substituting t = 5:
x = 15(5)^2 - 2.0(5)^3
x = 375 - 250
x = 125
Therefore, the maximum positive coordinate reached by the particle is 125 meters, and it is reached at t = 5 seconds.
(c) To find the maximum positive velocity reached by the particle, we need to find the maximum value of v. To do this, we can find the derivative of v with respect to t and set it equal to zero:
a = dv/dt = d(30t - 6.0t^2)/dt
a = 30 - 12.0t
Setting a = 0 and solving for t:
30 - 12.0t = 0
12.0t = 30
t = 2.5 s
To determine the maximum positive velocity, we substitute t = 2.5 into the velocity equation:
v = 30t - 6.0t^2
v = 30(2.5) - 6.0(2.5)^2
v = 75 - 37.5
v = 37.5 m/s
Therefore, the maximum positive velocity reached by the particle is 37.5 m/s, and it is reached at t = 2.5 s.
(d) At the instant the particle is not moving (other than at t = 0), its velocity is zero. To find the acceleration at this instant, we substitute v = 0 into the acceleration equation:
a = 30 - 12.0t
0 = 30 - 12.0t
12.0t = 30
t = 2.5 s
Therefore, the acceleration of the particle at the instant it is not moving (other than at t = 0) is 30 - 12.0(2.5) = -15 m/s^2.
(e) To determine the average velocity of the particle between t = 0 and t = 3 s, we use the formula:
average velocity = (change in position) / (change in time)
Substituting the given values:
average velocity = (x_final - x_initial) / (t_final - t_initial)
average velocity = (81 - 0) / (3 - 0)
average velocity = 81 / 3
average velocity = 27 m/s
Therefore, the average velocity of the particle between t = 0 and t = 3 s is 27 m/s.
To answer these questions, we need to use the given position function x = 15t^2 - 2.0t^3.
(a) To determine the position, velocity, and acceleration of the particle at t = 3.0 s, we substitute t = 3.0 into the position equation:
x = 15(3.0)^2 - 2.0(3.0)^3
x = 135 - 54
x = 81 meters
So, at t = 3.0 s, the position of the particle is 81 meters.
To find the velocity, we need to find the derivative of the position function with respect to time:
v = d/dt (15t^2 - 2.0t^3)
v = 30t - 6.0t^2
Substituting t = 3.0 into the velocity equation:
v = 30(3.0) - 6.0(3.0)^2
v = 90 - 54
v = 36 m/s
So, at t = 3.0 s, the velocity of the particle is 36 m/s.
To find the acceleration, we take the derivative of the velocity function with respect to time:
a = d/dt (30t - 6.0t^2)
a = 30 - 12.0t
Substituting t = 3.0 into the acceleration equation:
a = 30 - 12.0(3.0)
a = 30 - 36
a = -6 m/s^2
So, at t = 3.0 s, the acceleration of the particle is -6 m/s^2.
(b) To find the maximum positive coordinate reached by the particle, we need to find the maximum value of the position function. To do this, we can take the derivative of the position function and set it equal to zero:
dx/dt = d/dt (15t^2 - 2.0t^3)
dx/dt = 30t - 6.0t^2
0 = 30t - 6.0t^2
Solving this equation, we find two possible values for t: t = 0 and t = 5.
Since t = 0 corresponds to the starting position, we can ignore it. Thus, the maximum positive coordinate is reached when t = 5. Substituting t = 5 into the position function:
x = 15(5)^2 - 2.0(5)^3
x = 375 - 250
x = 125 meters
So, the maximum positive coordinate reached by the particle is 125 meters, and it occurs at t = 5 seconds.
(c) To find the maximum positive velocity reached by the particle, we need to find the maximum value of the velocity function. To do this, we can take the derivative of the velocity function and set it equal to zero:
dv/dt = d/dt (30t - 6.0t^2)
dv/dt = 30 - 12.0t
0 = 30 - 12.0t
Solving this equation, we find t = 2.5. Substituting t = 2.5 into the velocity function:
v = 30(2.5) - 6.0(2.5)^2
v = 75 - 37.5
v = 37.5 m/s
So, the maximum positive velocity reached by the particle is 37.5 m/s, and it occurs at t = 2.5 seconds.
(d) The acceleration of the particle at the instant the particle is not moving (other than at t = 0) can be found by substituting t = 3 into the acceleration function:
a = 30 - 12.0(3)
a = 30 - 36
a = -6 m/s^2
So, the acceleration of the particle at the instant the particle is not moving is -6 m/s^2.
(e) The average velocity of the particle between t = 0 and t = 3 can be found by calculating the displacement and dividing it by the time:
Displacement = x(3) - x(0)
Displacement = 15(3)^2 - 2(3)^3 - (15(0)^2 - 2(0)^3)
Displacement = 135 - 54 - (0 - 0)
Displacement = 81 meters
Average velocity = Displacement / Time
Average velocity = 81 meters / 3 seconds
Average velocity = 27 m/s
So, the average velocity of the particle between t = 0 and t = 3 seconds is 27 m/s.
x = 15t^2 - 2.0t^3
v = dx/dt = 30 t - 6 t^2
a = d^2x/dt^2 = 30 - 12 t
part a, just put in 3 for t in the above
part b
max position when v = 0
0 = t (30 - 6t)
so at t = 5 sec
then
x = 15*25 - 2*125 = 125
part c
max v when a = 0
t = 30/12
v = 30 (30/12) - 6 (30/12)^2
part d
not moving at t = 5 from part b
a = 30 - 12(5)
part e
Vmean (3) = x at 3 - x at 0
Vmean = [15(9) - 2.0(27) - 0 ]/3