Question

A particle moves along the path y=x^3+3x+1 where units are in centimetres.If the horizontal velocity Vx is constant at 2cm/s,find the magnitude and direction of the velocity of the particle at the point (1,5).
dy/dt = 3x^2+3 dx/dt
so, since Vx=dx/dt=2,
dy/dt = 6(x^2+1)
at x=1, Vy=dy/dt = 12

so, V=√(4+144)
tanθ = Vy/Vx = 6

Related Questions

At t = 0, a particle leaves the origin with a velocity of 9 m/s in the positive y direction and move... A body starts with an initial velocity of 10m/s and moves along a straight line with a constant acce... Context: The function v(t) represents the velocity of a particle moving along a horizontal line at a... if a particle moves along a horizontal path with an acceleration of a=(2t-9) m/s2 ( t is in seconds)...