A) At what time in the future will the x-component of the mass’s position be 0m?

Question

A) At what time in the future will the x-component of the mass’s position be 0m?

B) At the time when the x-component of the mass’s position is 0m, what is the y-component of position?

C) What is the velocity at the time when the x-component of the mass’s position is 0m?

D) The position of a particle is given by? r(t)=Acos(?t+?)ˆ?+Asin(?t+?)jˆ.

The variables A, ?, and ? are positive constants and ? is between 0

and ?/2 . At what time t0 is ?r(t0) = ?Aˆ? + Ajˆ?

A mass is moving with constant acceleration in the x-y plane. At time t-0s the position is 4mit6mj, it is moving at a speed of 10% at an angle of 60° with the x-axis and 30° with the y-axis, and it is subject to an acceleration of-4 m A

Explanation / Answer

As given in the question,

initial position = (4, 6)

Speed: u(x) = 10 cos 60° = 5 m/s and u(y) = 10 cos 30° = 8.66 m/s

Acceleration: a(x) = - 4 m/s^2 and a(y) = 0

(a) For the time in the future will the x-component of the mass’s position be 0,

  s = u*t + (1/2)*a*t^2

  => 0-4 = u(x)*t + (1/2)*a(x)*t^2

  => - 4 = 5*t + (1/2)*(-4)*t^2

  => 2*t^2 - 5*t - 4 = 0 =>   t = 3.14 s

(b) For the y-component of position at this time,

Since a(y) = 0, s(y) = u(y)*t  = 8.66*3.14 = 27.19 m

So, the y-component of final position = 6 + 27.19 = 33.19 m

(c) For the velocity at the time when the x-component of the mass’s position is 0,

v(x) = u(x) + a(x)*t  = 5 + (-4)*3.14 = - 7.56 m/s

v(y) = u(y) = 8.66 m/s

So, the velocity at this position = v(x) i^ + v(y) j^ = - 7.56 i^ + 8.66 j^

(d) Supoose for time t0, r(t0) = A i^ + A j^

For y-directional position, A = u(y)*t0 =>   A = 8.66*t0

For x-directional position, - A = u(x)*t0 + (1/2)*a(x)*t0^2

=> - A = 5*t0 + (1/2)*(-4)*t0^2

=> - A = 5*t0 - 2*t0^2

putting the value of A in terms of t0 from y-directional motion,

- 8.66*t0 = 5*t0 - 2*t0^2

=> 2*t0 = 5 + 8.66 = 13.66

=>   t0 = 13.66/2 = 6.83 s

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