A particle which moves with curvilinear motion has coordinates in millimeters which vary
with the time t in seconds according to and Determine
the magnitudes of the velocity v and acceleration a and the angles which these vectors make
with the x-axis when t=0.4I sec
Given the coordinates in millimeters of the particle as a function of time t:
x(t) = 6t^2 - 4t + 2
y(t) = 3t^2 + 2t
To determine the velocity and acceleration vectors at t = 0.4 sec, we need to find the derivatives of the position vector with respect to time:
r(t) = xi + yj
v(t) = dr/dt = (dx/dt)i + (dy/dt)j
a(t) = dv/dt = (d^2x/dt^2)i + (d^2y/dt^2)j
First, calculate the position vector at t = 0.4 sec:
x(0.4) = 6(0.4)^2 - 4(0.4) + 2 = 1.6
y(0.4) = 3(0.4)^2 + 2(0.4) = 1.52
r(0.4) = 1.6i + 1.52j
Now, differentiate to find the velocity vector:
v(0.4) = (d/dt)(1.6)i + (d/dt)(1.52)j
v(0.4) = (12t - 4)i + (6t + 2)j
v(0.4) = (4.8)i + (4.4)j
Calculate the magnitude of the velocity vector:
|v(0.4)| = sqrt[(4.8)^2 + (4.4)^2] = sqrt[23.04 + 19.36] = sqrt[42.4] ≈ 6.51 mm/s
Calculate the angle with the x-axis:
tanθ = (4.4)/(4.8) = 0.9167
θ ≈ 43.55°
Now, differentiate to find the acceleration vector:
a(0.4) = (d^2x/dt^2)i + (d^2y/dt^2)j
a(0.4) = 12i + 6j
Calculate the magnitude of the acceleration vector:
|a(0.4)| = sqrt[(12)^2 + (6)^2] = sqrt[144 + 36] = sqrt[180] = 13.42 mm/s^2
Calculate the angle with the x-axis:
tanθ = 6/12 = 0.5
θ ≈ 26.57°
Therefore, at t = 0.4 sec, the magnitude of the velocity vector is approximately 6.51 mm/s making an angle of 43.55° with the x-axis, and the magnitude of the acceleration vector is approximately 13.42 mm/s^2 making an angle of 26.57° with the x-axis.