The velocity components in x and y of a particle moving in curvilinear motion in

Question

The velocity components in x and y of a particle moving in curvilinear motion in the cartesian plane are: V_x = 3t^2 + 1 V_y = t^3 + t Where t is in second and the velocities are in m/s a) Determine the magnitude of the velocity when 2 seconds has passed. b) Determine the magnitude of the acceleration when 2 seconds has passed. c) If the particle is initially (t = 0) at the origin (x = 0, and y = 0) what is its position when 2 seconds has passed. d) It is known that the radius of curvature at the instant of t = 2 s is rho = 82.4 m. What is the normal acceleration component? e) What is the tangential acceleration component? Instead of trying to calculate the tangential acceleration using the derivative of the speed you can just use the result of part b) and the definition of the magnitude of the acceleration based on the normal and tangential components.

Explanation / Answer

part a:

at t=2 seconds, Vx=3*t^2+1=13 m/s

at t=2 seconds, Vy=t^3+t=10 m/s

so magnitude of velocity=sqrt(Vx^2+Vy^2)

=sqrt(13^2+10^2)

=16.4 m/s

part b:

acceleration along x axis=Ax=dVx/dt=6*t

acceleration along y axis=Ay=dVy/dt=3*t^2+1

so at t=2 seconds, Ax=12 m/s^2

at t=2 seconds,Ay=13 m/s^2

magnitude of acceleration=sqrt(12^2+13^2)=17.692 m/s^2

part c:

as Vx=dx/dt

==>dx/dt=3*t^2+1

==>x=integration of (3*t^2+1)*dt=t^3+t+c

where c is a constant

at t=0, x=0

hence c=0

so x=t^3+t

similarly Vy=dy/dt

==>y=integration of (t^3+t)*dt

==>y=0.25*t^4+0.5*t^2

at t=2 seconds, x=10 m
at t=2 seconds, y=6 m

hence position is x=10 m, y=6 m

part d:
normal acceleration of the component=magnitude of speed^2/radius of curvature

=16.4^2/82.4=3.2641 m/s^2

part e:

as total acceleration=sqrt(normal acceleration^2+tangential acceleration^2)

==>17.692=sqrt(3.2641^2+tangential acceleration^2)

==>tangential acceleration=sqrt(17.692^2-3.2641^2)=17.388 m/s^2

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