6. 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=0.4*041 sec
Given:
x = 3t^2 - 2t^3
y = 2t^2 - 3t^3
Differentiating x and y with respect to time t:
vx = d(x)/dt = 6t - 6t^2
vy = d(y)/dt = 4t - 9t^2
When t=0.4*041 sec:
t = 0.4*041 = 0.1644 sec
Substitute t=0.1644 sec into the equations for vx and vy:
vx = 6(0.1644) - 6(0.1644)^2
vx = 0.9864 - 0.096766
vx = 0.8896 mm/s
vy = 4(0.1644) - 9(0.1644)^2
vy = 0.6576 - 0.24167616
vy = 0.4159 mm/s
Calculate the magnitude of velocity:
v = sqrt(vx^2 + vy^2)
v = sqrt(0.8896^2 + 0.4159^2)
v = sqrt(0.7909424 + 0.17320881)
v = sqrt(0.96415121)
v = 0.98 mm/s
Calculate the angle between the velocity vector and the x-axis:
tanθ = vy / vx
θ = arctan(vy / vx)
θ = arctan(0.4159 / 0.8896)
θ ≈ 25.14°
Differentiating vx and vy with respect to time t again to find acceleration components:
ax = d^2(x)/dt^2 = 6 - 12t
ay = d^2(y)/dt^2 = 4 - 18t
Substitute t=0.1644 sec into the equations for ax and ay:
ax = 6 - 12(0.1644)
ax = 6 - 1.9728
ax = 4.0272 mm/s^2
ay = 4 - 18(0.1644)
ay = 4 - 2.9572
ay = 1.0428 mm/s^2
Calculate the magnitude of acceleration:
a = sqrt(ax^2 + ay^2)
a = sqrt(4.0272^2 + 1.0428^2)
a = sqrt(16.225 + 1.08729)
a = sqrt(17.31229)
a = 4.16 mm/s^2
Calculate the angle between the acceleration vector and the x-axis:
tanφ = ay / ax
φ = arctan(ay / ax)
φ = arctan(1.0428 / 4.0272)
φ ≈ 14.82°
Therefore, when t = 0.4*041 sec:
- The magnitude of the velocity v is 0.98 mm/s at an angle of approximately 25.14° with the x-axis.
- The magnitude of the acceleration a is 4.16 mm/s^2 at an angle of approximately 14.82° with the x-axis.