A particle has a constant acceleration of = (5.5 m/s2) + (3.7 m/s2) . At time t = 0, the velocity is zero and the position vector is 0 = (9.8 m).
(a) Find the velocity and position vectors as functions of time t.
(b) Find the equation of the particle's path in the xy plane.
To find the velocity vector as a function of time, we can use the kinematic equation:
v = v0 + at
Given that the initial velocity v0 is zero, and the constant acceleration vector a = (5.5 m/s^2) + (3.7 m/s^2), we can substitute these values into the equation:
v = 0 + (5.5 m/s^2 + 3.7 m/s^2) * t
Simplifying the equation, we get:
v = (9.2 m/s) * t
To find the position vector as a function of time, we need to integrate the velocity equation:
r = ∫(v) dt
Integrating the equation, we get:
r = ∫(9.2 m/s) * t dt
Integrating with respect to t gives:
r = (9.2 m/s) * (t^2)/2 + C
Where C is the constant of integration.
Given that the position vector at t = 0 is 0 = (9.8 m), we can substitute these values into the equation to solve for the constant C:
0 = (9.2 m/s) * (0^2)/2 + C
C = 0
Therefore, the equation for the position vector as a function of time is:
r = (9.2 m/s) * (t^2)/2
For part (b), to find the equation of the particle's path in the xy plane, we need to express the position vector in terms of x and y coordinates.
Given that the position vector is r = (9.2 m/s) * (t^2)/2, we can express it in terms of x and y coordinates by substituting:
x = (9.2 m/s) * (t^2)/2
y = 0
Therefore, the equation of the particle's path in the xy plane is:
y = 0
This indicates that the particle only moves in the x-direction and does not move in the y-direction. Thus, its path is a straight line on the x-axis.