At t=0 a particle starts from rest at x=0, y=0 and moves in the xy plane with an acceleration a=(4.0i+3.0j)m/s^2. Determine (a) the x and y components of velocity, (b) the speed of the particle, and (c) the position of the particle, all as a function of time. (d) evaluate all the above at t=2.0s.
QUESTION: Assume that you are supplied with a translational potentiometer sensor with the following properties;
* 100 V DC Power Supply,
* 1 resistive wire: 40 mm long with 1mm2 cross-section area. ( resistivity=0,00025 Ohm m)
It is also given to you that;
-In the initial state (where no force is applied to the system) of the potentiometer, movable arm of the output voltage positioner covers 90% (length) of the resistive wire.
- When 10N of force is applied to the system you observe 6mm displacement (in (-) y direction) of the movable arm of the output voltage positioner
Calculate what is the output voltage change caused by the 10N of force?
To determine the x and y components of velocity, we can integrate the given acceleration function with respect to time.
(a) Integrating the acceleration function yields the velocity function:
v = ∫(a dt)
v = ∫(4.0i + 3.0j) dt
v = (4.0t)i + (3.0t)j + C
Here, C is the constant of integration. Since the particle starts from rest at t=0, the initial velocity is zero.
Plugging in t=0, we have:
0 = (4.0*0)i + (3.0*0)j + C
0 = 0i + 0j + C
C = 0
Therefore, the velocity function becomes:
v = (4.0t)i + (3.0t)j
(b) To find the speed of the particle, we can use the magnitude of the velocity vector:
speed = |v|
speed = |(4.0t)i + (3.0t)j|
speed = √((4.0t)^2 + (3.0t)^2)
speed = √(16t^2 + 9t^2)
speed = √(25t^2)
speed = 5t
(c) To determine the position of the particle as a function of time, we need to integrate the velocity function:
r = ∫(v dt)
r = ∫((4.0t)i + (3.0t)j) dt
r = (2.0t^2)i + (1.5t^2)j + C1
Here, C1 is the constant of integration. To find the position at t=0, we have:
r = (2.0*0^2)i + (1.5*0^2)j + C1
r = 0i + 0j + C1
r = C1
Therefore, the position function becomes:
r = (2.0t^2)i + (1.5t^2)j + C1
(d) To evaluate all the above at t=2.0s, we substitute t=2.0 into the velocity and position functions:
(a) The x and y components of velocity at t=2.0s are:
v = (4.0 * 2.0)i + (3.0 * 2.0)j
v = 8.0i + 6.0j
(b) The speed of the particle at t=2.0s is:
speed = 5 * 2.0
speed = 10.0 m/s
(c) The position of the particle at t=2.0s is:
r = (2.0 * 2.0^2)i + (1.5 * 2.0^2)j + C1
r = 8.0i + 6.0j + C1
Note that the value of C1 is not given, so we cannot determine the specific position without that information.
To determine the x and y components of velocity, we integrate the given acceleration function with respect to time. Since the particle starts from rest, the initial velocities in both x and y directions will be zero.
Given acceleration, a = (4.0i + 3.0j) m/s^2
(a) To find the x and y components of velocity, integrate the acceleration function with respect to time:
Integrating a_x = 4.0 m/s^2 with respect to t gives:
v_x = ∫(4.0 dt) = 4.0t + C1
Integrating a_y = 3.0 m/s^2 with respect to t gives:
v_y = ∫(3.0 dt) = 3.0t + C2
Since the initial velocities are both zero (particle starts from rest), C1 = C2 = 0.
Therefore, the x and y components of velocity are:
v_x = 4.0t
v_y = 3.0t
(b) The speed of the particle, given by the magnitude of the velocity vector, is determined by calculating the magnitude of the velocity vector:
speed = |v| = sqrt(v_x^2 + v_y^2)
Substituting the values of v_x and v_y obtained above:
speed = sqrt((4.0t)^2 + (3.0t)^2) = sqrt(16.0t^2 + 9.0t^2) = sqrt(25.0t^2) = 5.0t
Therefore, the speed of the particle is given by speed = 5.0t.
(c) To find the position of the particle as a function of time, we integrate the x and y components of the velocity with respect to time.
Integrating v_x = 4.0t dt gives:
x = ∫(4.0t dt) = 2.0t^2 + C3
Integrating v_y = 3.0t dt gives:
y = ∫(3.0t dt) = 1.5t^2 + C4
Since the particle starts at (x,y) = (0,0), we know that (0,0) satisfies the equations above. So, C3 = C4 = 0.
Therefore, the position of the particle as a function of time is:
x = 2.0t^2
y = 1.5t^2
(d) To evaluate the values at t = 2.0s, substitute t = 2.0 into the expressions obtained in parts (a), (b), and (c):
(a) x and y components of velocity:
v_x = 4.0 * 2.0 = 8.0 m/s
v_y = 3.0 * 2.0 = 6.0 m/s
(b) Speed of the particle:
speed = 5.0 * 2.0 = 10.0 m/s
(c) Position of the particle:
x = 2.0 * (2.0)^2 = 8.0 m
y = 1.5 * (2.0)^2 = 6.0 m
Therefore, at t = 2.0s:
(a) The x and y components of velocity are 8.0 m/s and 6.0 m/s, respectively.
(b) The speed of the particle is 10.0 m/s.
(c) The position of the particle is (8.0, 6.0) m.
a) 4t
b)3t