Question

Velocity in xy-Plane position in the xy plane is given by the vector (ct 4dtt)i +(2ct? -d')j, where c and d are positive c FInd the expression for the x-component of the velocity (for time t > 0) when the in the z-direction. You should express your answer in terms of the variables c and d. Find the n for the y component of the velocity (for timet 0) when the particle is moving in the y-direction. D

Answer 1

Solving the problem using the definition of instantaneous velocity.

Part(A) given, position of particle, r bar = (ct^2 - 4dt^3) i^+ (2ct^2 - dt^2) j^so, we cab say, r = ct^2 - 4dt^3 & r_y = 2ct^2 - dt^3 r bar = (r_2) i^+ (r_y)j^Also, by definition we know that, if 'r' is the position of any particular any instant of time = 't' then, it's instantaneous velocity V is given by V = dr/dt = small change in position/small change in time so, for the given scenario, the velocity of the particle can be represented as V bar = d/dt (r bar) = d/dt (r_x) i^+ d/dt (r_y) j^V bar = V_x (i^) + V_y j^so, V_x = d/dt (r_2) & V_y = d/dt (r_y) so, for time, (t > 0) 2 component of velocity 0 V_x = d/dt (r_2) = d/dt (ct^2 - 4dt^3) = 2ct - 12dt^2 so, x component of velocity is V_x = 2ct - 12dt^2 along i^rightdoublearrow part (B) leftdoublearrow similarly y component of velocity is V_y = d(r_y)/dt = d/dt (2ct^2 - dt^3)