A particle moves in the xy plane with a constant acceleration given by a = -4.0j m/s^2. At t = 0, its position and velocity are 10i m and (-2.0i +8.0j) m/s, respectively. What is the distance from the origin to the particle at t = 2.0 s?
Oh, so we have a particle that is constantly accelerating. It's like that friend who can't sit still and is always on the move! Let's find out where this little troublemaker ends up.
We know that acceleration is given by a = -4.0j m/s^2. Now, I'm no math whiz, but that negative sign tells me that this particle is moving downward in the y-direction. Maybe it's trying to find buried treasure down there!
At t = 0, the position of the particle is 10i m. I like to think of this as the starting point for a wild roller coaster ride. Buckle up!
Now, the velocity of the particle at t = 0 is (-2.0i +8.0j) m/s. That means it's moving to the left in the x-direction (-2.0i), but also moving upward in the y-direction (8.0j). This particle can't make up its mind on which way it wants to go!
So, how do we find the distance from the origin to the particle at t = 2.0 s? Well, we can use the following equation:
s = ut + 0.5at^2
Where s is the displacement, u is the initial velocity, t is the time, and a is the constant acceleration.
Since the particle starts at the origin, its initial displacement is 0. Therefore, we only need to consider the second part of the equation.
s = 0.5at^2
We plug in the values we have:
s = 0.5(-4.0j)(2.0s)^2
Calculating that gives us:
s = 0.5(-4.0j)(4.0s^2)
s = -8.0j(4.0s^2)
Uh-oh, it looks like we have a mixed-up particle! It's still moving downward in the y-direction, but now it's travelling at double the speed. It's really in a rush to find that buried treasure!
Now, we just need to calculate the magnitude of the displacement vector to find the distance from the origin. In other words, we need to find the absolute value of the y-component of the displacement vector:
|s| = |-8.0j|*(4.0s^2)
|s| = 8.0*(4.0s^2)
|s| = 32.0s^2
Now, we plug in the value of t = 2.0 s:
|s| = 32.0*(4.0(2.0)^2)
|s| = 32.0*(4.0(4.0))
|s| = 32.0*(16.0)
|s| = 512.0
Voilà! The distance from the origin to the particle at t = 2.0 s is 512.0 meters. That particle certainly knows how to put on a show!
To find the distance from the origin to the particle at t = 2.0 s, we need to determine the position of the particle at that time.
Given:
Initial position, r(0) = 10i m
Initial velocity, v(0) = (-2.0i + 8.0j) m/s
Acceleration, a = -4.0j m/s^2
To find the position of the particle at t = 2.0 s, we can use the kinematic equation:
r(t) = r(0) + v(0)t + (1/2)at^2
Substituting the given values, we have:
r(2.0) = 10i + (-2.0i + 8.0j)(2.0) + (1/2)(-4.0j)(2.0)^2
= 10i - 4.0i + 16.0j - 4.0j
= 6.0i + 12.0j
The position vector, r(2.0), is 6.0i + 12.0j.
To find the distance from the origin to this position, we can use the Pythagorean theorem:
distance = √(x^2 + y^2)
where x is the x-component of the position vector and y is the y-component.
distance = √((6.0)^2 + (12.0)^2)
= √(36 + 144)
= √180
= 6√5
Therefore, the distance from the origin to the particle at t = 2.0 s is 6√5 meters.
To find the distance from the origin to the particle at t = 2.0 s, we need to find the particle's position at that time. We are given its initial position and velocity, as well as the acceleration.
First, let's find the particle's position at t = 2.0 s using the kinematic equations:
r(t) = r(0) + v(0)t + (1/2)a t^2
Using the given values, we have:
r(0) = 10i m (initial position)
v(0) = (-2.0i + 8.0j) m/s (initial velocity)
a = -4.0j m/s^2 (acceleration)
t = 2.0 s (time)
Substituting these values into the kinematic equation, we get:
r(t) = 10i m + (-2.0i + 8.0j) m/s * 2.0 s + (1/2) * (-4.0j m/s^2) * (2.0 s)^2
Calculating each term separately:
r(t) = 10i m + (-2.0i + 8.0j) m/s * 2.0 s + (1/2) * (-4.0j m/s^2) * 4.0 s^2
= 10i m + (-4.0i + 16.0j) m + (-2.0j m/s^2) * 16.0 s^2
= (10i - 4.0i) m + (16.0j - 32.0j) m
= 6.0i m - 16.0j m
Now, we can find the distance from the origin to the particle at t = 2.0 s, which is given by the magnitude of the position vector:
distance = |r(t)|
= |6.0i m - 16.0j m|
= sqrt((6.0)^2 + (-16.0)^2) m
= sqrt(36.0 + 256.0) m
= sqrt(292.0) m
≈ 17.1 m
Therefore, the distance from the origin to the particle at t = 2.0 s is approximately 17.1 meters.
a(t) = 0i - 4.0j
v(t) = ci - 4t+dj
v(0) = -2i+8j so c=-2 and d=8
so, v(t) = -2i + (8-4t)j
s(t) = (-2t+c)i + (8t-2t^2+d)j
s(0) = 10i, so c=10 and d=0
s(t) = (10-2t)i + (8t-2t^2)j
|s(2)| = |6i+8j| = 10