Physics Question

Question

The acceleration of a particle moving only on a horizontal xy plane is given by ?→=5??ˆ+6??ˆa→=5ti^+6tj^, where ?→a→ is in meters per second-squared and t is in seconds. At t = 0, the position vector ?→=(26.0m)?ˆ+(36.0m)?ˆr→=(26.0m)i^+(36.0m)j^ locates the paticle, which then has the velocity vector ?→=(6.70m/s)?ˆ+(1.90m/s)?ˆv→=(6.70m/s)i^+(1.90m/s)j^. At t = 4.60 s, what are (a) its position vector in unit-vector notation and (b)the angle between its direction of travel and the positive direction of the x axis?

Answer 1

So Ashton F.,

Since the acceleration is nicely orthogonalized into i and j components, just integrate each of those separately, the first time to get expression for delta v(t) components, the second time to get delta r(t) components. Don't forget to include the initial conditions, at the appropriate places. The vector for v (t=4.6 sec) makes an angle with the x-axis, you have the components of the tangent, obviously.

-- Cheers, -- Mr. d.