A particle has ⃗r(0) = (4 m)j and ⃗v(0) = (2 m/s)i.
If its acceleration is constant and given by ⃗a = −(2 m/s2) (i +j), at what time t does the particle first cross the x axis?
Answer in units of s.
part 2. At what time t is the particle moving parallel to the y axis; that is, in the j direction?
Answer in units of s.
Let u = x direction velocity (i) and v = y direction (j) velocity
Xo = 0
Yo = 4
Uo = +2
Vo = 0
Ax = -2
Ay = -2
u = Uo + Ax t
u = 2 - 2 t
x = Xo + Uo t +(1/2) Ax t^2
x = 2 t - t^2
y = Yo + Vo t + (1/2) Ay t^2
y = 4 - t^2
so
y = 0 when t = 2
u = 0 when t = 1
Thank you very much.. i understand it...it crosses the x-axis when the y point is =o, and is parallel to the y- axis when the vector in the x direction aka the slope is = to 0 =]
To find the time at which the particle first crosses the x-axis, we need to determine when the x-component of its position becomes zero.
Step 1: Find the x-coordinate equation.
The particle's position ⃗r as a function of time t can be expressed as:
⃗r(t) = ⃗r(0) + ⃗v(0)t + 0.5⃗at^2
Since ⃗a = −(2 m/s^2)(i + j), the equation becomes:
⃗r(t) = (4 m)j + (2 m/s)i * t + 0.5(-2 m/s^2)(i + j)t^2
Breaking it down into x and y components, we have:
x(t) = (2 m/s)t - (m/s^2)t^2
y(t) = 4 m + (2 m/s)t - (m/s^2)t^2
Step 2: Set x(t) = 0 and solve for t.
To find the time at which the particle crosses the x-axis, set x(t) = 0 and solve for t:
0 = (2 m/s)t - (m/s^2)t^2
Rearranging the equation, we get:
(m/s^2)t^2 - (2 m/s)t = 0
Factoring out t, we have:
t(t - 2 s) = 0
Setting each factor equal to zero, we get:
t = 0 s (This is the initial time when the particle is at position ⃗r(0))
t - 2 s = 0
Solving for t, we find:
t = 2 s
Therefore, the particle first crosses the x-axis at t = 2 seconds.
Now let's move on to part 2.
To find the time at which the particle is moving parallel to the y-axis (in the j direction), we need to determine when the x-component of its velocity becomes zero.
Step 1: Find the x-component of velocity.
The particle's velocity ⃗v as a function of time t can be expressed as:
⃗v(t) = ⃗v(0) + ⃗at
Since ⃗a = −(2 m/s^2)(i + j), the equation becomes:
⃗v(t) = (2 m/s)i + (-2 m/s^2)(i + j)t
Breaking it down into x and y components, we have:
v_x(t) = (2 m/s) - (2 m/s^2)t
v_y(t) = 0 m/s - (2 m/s^2)t
Step 2: Set v_x(t) = 0 and solve for t.
To find the time at which the x-component of velocity becomes zero, set v_x(t) = 0 and solve for t:
0 = (2 m/s) - (2 m/s^2)t
Rearranging the equation, we get:
(2 m/s^2)t = (2 m/s)
Simplifying, we find:
t = 1 s
Therefore, the particle is moving parallel to the y-axis at t = 1 second.
To find the time at which the particle first crosses the x-axis, we need to determine when the y-coordinate of the particle becomes zero.
The equation for the y-coordinate of the particle as a function of time t is given by:
y(t) = y0 + v0yt + (1/2)ayt^2
where y0 is the initial y-coordinate, v0y is the initial y-component of velocity, ay is the y-component of acceleration, and t is the time.
In this case, y0 = 4 m, v0y = 0 (since the particle's initial velocity is purely in the x-direction), and ay = -2 m/s^2.
Substituting these values into the equation, we have:
y(t) = 4 m + 0 + (1/2)(-2 m/s^2)t^2
= 4 m - (1 m/s^2)t^2
To find when the particle crosses the x-axis, we need to solve for t when y(t) = 0:
4 m - (1 m/s^2)t^2 = 0
Now we can solve for t by rearranging the equation:
(1 m/s^2)t^2 = 4 m
t^2 = 4 s^2
t = √(4 s^2)
t = 2 s
Therefore, the particle first crosses the x-axis at t = 2 seconds.
Part 2:
To find the time at which the particle is moving parallel to the y-axis (in the j direction), we need to determine when the x-component of velocity becomes zero.
The equation for the x-component of velocity is given by:
vx(t) = v0x + axt
In this case, v0x = 2 m/s and ax = -2 m/s^2 (since the acceleration is in the negative x-direction).
Substituting these values into the equation, we have:
vx(t) = 2 m/s + (-2 m/s^2) t
To find when the particle is moving parallel to the y-axis, vx(t) = 0:
2 m/s + (-2 m/s^2) t = 0
Now we can solve for t by rearranging the equation:
(-2 m/s^2)t = -2 m/s
t = -2 m/s / (-2 m/s^2)
t = 1 s
Therefore, the particle is moving parallel to the y-axis (in the j direction) at t = 1 second.