A particle leaves the origin with a velocity of 7.2 m/s in the positive y direction and moves in the xy plane with a constant acceleration of (3.0 – 2.0 ˆi ˆj ) m/s2. At the instant the particle moves back across the x axis (y = 0), what is the value of its x coordinate?
susan drops a ball, and 2 second later the ball has a speed of 20m/s. what is the ball's acceleration?
Well, well, well, looks like our particle has taken a detour back across the x-axis. Naughty particle! Let's find out its x-coordinate, shall we?
We know that the particle has a constant acceleration of (3.0 – 2.0 ˆi ˆj ) m/s2. If we break it down, we see that it has an acceleration of 3.0 m/s² in the x-direction and -2.0 m/s² in the y-direction. That's one fancy particle!
Since the acceleration is constant, we can employ some good ol' kinematic equations to solve this problem. But before we do that, let's take a moment to appreciate the particle's sense of direction. Going back and forth across the x-axis like it's on a cosmic rollercoaster ride! Whee!
To find the x-coordinate when the particle crosses the x-axis (y = 0), we need to determine the time it takes for that to happen. Let's say it's t seconds.
The initial velocity in the y-direction is given as 7.2 m/s, and since the acceleration in that direction is -2.0 m/s², we can use the equation:
0 = 7.2 - 2.0t
Solving for t, we get t = 7.2/2.0, which is 3.6 seconds. Yay, math!
Now that we know the time, we can determine the x-coordinate using the x-direction's constant acceleration. The initial velocity in the x-direction is zero and the acceleration is 3.0 m/s². Again, we can use a kinematic equation, this time:
x = 0.5at²
Substituting in the values we know, we get:
x = 0.5(3.0)(3.6)²
Crunching the numbers gives us x = 19.44 meters. Ta-da!
So, the particle's x-coordinate when it crosses the x-axis (y = 0) is 19.44 meters. That particle sure knows how to make a grand entrance, doesn't it? Keep those physics questions coming, and I'll keep doing my best to entertain you with my "clown"-ishing explanations!
To find the value of the x-coordinate when the particle crosses the x-axis (y = 0), we need to determine the time it takes for the particle to reach y = 0.
Given:
Initial velocity (u) = 7.2 m/s in the positive y-direction
Acceleration (a) = (3.0 - 2.0 ˆi ˆj) m/s^2
Since the acceleration is constant, we can use the kinematic equation to find the time (t) taken for the particle to reach y = 0.
The kinematic equation for motion in the y-direction is given by:
y = ut + (1/2)at^2
Since the initial position y0 is 0 (particle starts from the origin), the equation becomes:
0 = (7.2)t + (1/2)(3.0 - 2.0)t^2
Simplifying the equation gives:
0 = (7.2)t + (1.5 - t^2)
Rearranging the equation gives a quadratic equation:
t^2 - (7.2)t + 1.5 = 0
Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -7.2, and c = 1.5.
Solving the quadratic equation using the quadratic formula gives two solutions: t ≈ 7.015 and t ≈ 0.215.
Since the time cannot be negative, we discard t ≈ 7.015. Therefore, the value of t when the particle crosses the x-axis is t ≈ 0.215 seconds.
To find the x-coordinate of the particle at this time, we can use the equation:
x = ut + (1/2)at^2
Since the initial x-coordinate is also 0 (particle starts from the origin), the equation becomes:
x = (1/2)(3.0 - 2.0)(0.215)^2
Simplifying the equation gives:
x ≈ 0.0326 meters
Therefore, when the particle crosses the x-axis (y = 0), its x-coordinate is approximately 0.0326 meters.
To find the x-coordinate of the particle at the moment it moves back across the x-axis (y = 0), we need to determine the time it takes for the particle to reach y = 0 and then use that time to calculate the x-coordinate.
Let's break down the problem step by step:
Step 1: Find the time it takes for the particle to reach y = 0.
To find the time, we can use the kinematic equation:
y = y0 + v0y * t + 0.5 * ay * t^2
In this equation:
- y is the current y-coordinate (y = 0)
- y0 is the initial y-coordinate (y0 = 0)
- v0y is the initial velocity in the y-direction (v0y = 7.2 m/s)
- ay is the acceleration in the y-direction (ay = -2.0 m/s^2)
- t is the time we want to find
Since y0 and y are both zero, the equation simplifies to:
0 = 7.2 * t + 0.5 * (-2.0) * t^2
Now, we have a quadratic equation. Solving this equation for t will give us the time it takes for the particle to reach y = 0.
Step 2: Solve the quadratic equation for t.
The equation 0 = 7.2 * t + 0.5 * (-2.0) * t^2 can be rearranged as:
-2t^2 + 7.2t = 0
Factoring out t, we get:
t * (-2t + 7.2) = 0
This equation has two solutions:
t = 0 (which is the initial time) or -2t + 7.2 = 0
Solving the second equation:
-2t + 7.2 = 0
-2t = -7.2
t = -7.2 / -2
t = 3.6 seconds
The positive value for t, 3.6 seconds, is the time it takes for the particle to reach y = 0.
Step 3: Calculate the x-coordinate using the found time.
To find the x-coordinate, we can use the equation of motion:
x = x0 + v0x * t + 0.5 * ax * t^2
In this equation:
- x is the x-coordinate we want to find
- x0 is the initial x-coordinate (x0 = 0)
- v0x is the initial velocity in the x-direction (v0x = 0 since the particle starts at the origin)
- ax is the acceleration in the x-direction (ax = 3.0 m/s^2, given as (3.0 - 2.0))
Since v0x is zero, the equation simplifies to:
x = 0 + 0.5 * 3.0 * t^2
Substituting the value of t we found earlier (t = 3.6 seconds):
x = 0 + 0.5 * 3.0 * (3.6)^2
Evaluating this expression will give us the x-coordinate of the particle at the given time.