Kinematics/ angle/bearing?

A particle moves with constant acceleration and its velocity at time t :

v= (5-0.1t)i +(3+0.2t)j

1. find a, u

2. find the time t when particle is travelling in direction NE and find its speed.

3. find the distance and bearing of the particle from the starting point at that time.

2 Answers

Relevance
  • 1 month ago

    acceleration = dv/dt = a =  -0.1 *i  +0.2 j

    u = position = Integral(v dt) = (5t - 0.05t^2)*i + (3t + 0.1t^2)*j +u0  where u0 is a constant vector

    Define NE as 45deg north of east and let the i direction be pointing east, and the j diretion pointing north. THen

    45 = arctan(u_y/u_x)  or tan(u_y/u_x) = 1  -->  uy/u_x = pi/4 (radians)

    uy = pi*u_x/4 ---> 3t + 0.1t^2 = pi*(5t - 0.05t^2)/4

    re-arrange --> 3t + 0.1t^2 - pi*(5t - 0.05t^2)/4 = 0

    divide out a t --> 3 + 0.1t - pi*(5 - 0.05t)/4 = 0

    3-5*pi/4 +(0.1 +0.05*pi/4) t = 0 -->  t = (5*pi/4 - 3)/(0.1 +0.05*pi/4) 

    plug into your calculator for t.

    Speed is |v| = sqrt((5-0.1t)^2 +(3+0.2t)^2) using value of t found above

    For starting point - let starting point be at (0,0) --> u0 = 0 i + 0 j

    Distance is then  |u(t) - u0| = sqrt((5t - 0.05t^2)^2 +(3t + 0.1t^2)^2) evaluated at the value of t you found above

  • 1 month ago

    a=-0.1i+0.2j, u= v at t=0 =5i+3j,  NE is 45 degree. it is obvious. I strugle to find t in question 2. Thanks

Still have questions? Get your answers by asking now.