A particle has r(0)= (4m)j and v(0)= 92 m/s)i
If tts acceleration is constant and given by a=-(2 m/s^2)(i+j), at what time t does the particle first cross the x axis?
t=2 s, thanks Damon
Also,
at what time t is the partilce moving parallel to the y axis; that is the j direction?
That is when x'(t) = 0
from before: velocity =
r' = [-2 t + x'(0) ] i +[-2 t + y'(0)] j
= [ -2 t +92 ] i - 2 t j
so when -2 t + 92 = 0, there is no velocity component in the i direction
t = 46 seconds
thank you so much!!!
You are welcome.
To find the time at which the particle first crosses the x-axis, we need to determine the value of t when the position of the particle in the x-direction becomes zero (r_x = 0).
We are given that the initial position vector of the particle is r(0) = (4m)j, and the velocity vector is v(0) = 92 m/s)i. The acceleration of the particle is constant and given by a = -(2 m/s^2)(i+j).
Let's break down the problem into components:
1. Determine the position vector r(t) by integrating the velocity vector v(t):
r(t) = ∫v(t)dt
Given v(0) = 92 m/s)i, we can integrate the x-component to find r_x(t):
r_x(t) = ∫v_x(t)dt = ∫92 dt = 92t
2. Now, let's find the time when r_x(t) = 0 (crossing the x-axis):
92t = 0
Solving for t gives:
t = 0
So, the particle initially starts at the x-axis, at t = 0, and does not cross it at any later time.
Let's move on to the second part of the question: finding the time when the particle is moving parallel to the y-axis (j-direction).
To determine this time, we need to find when the x-component of the velocity vector v_x(t) becomes zero.
Given a = -(2 m/s^2)(i+j), the x-component of acceleration is a_x = -2 m/s^2.
From the equation of motion, we have:
v_x = u_x + a_xt
v_x = 92 - 2t
Now, let's find when v_x = 0:
92 - 2t = 0
Solving for t gives:
t = 46 seconds
So, the particle is moving parallel to the y-axis (j-direction) at t = 46 seconds.