3. At t = 0 , a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the xy plane with constant accelaration of ( 2.0i - 4.0j ) m/s2 . At the instant the x coordinate of the particle is 15 m , what is the speed of the particle ?

Well, it seems like our particle is having quite the adventure in the xy plane! Now, let's see if I can help you find the speed of the particle.

First, let's break down the motion of the particle into its x and y components. We are given that the particle has a constant acceleration of (2.0i - 4.0j) m/s^2. The x component of this acceleration is 2.0 m/s^2 and the y component is -4.0 m/s^2.

Now, we can find the time it takes for the particle to reach the x-coordinate of 15 m. The x-coordinate can be given by the equation x = x0 + v0xt + (1/2)at^2, where x0 is the initial x-coordinate, v0x is the initial velocity in the x-direction, t is the time, and a is the acceleration in the x-direction. Since the particle starts from the origin (x0 = 0) and has no initial x-velocity (v0x = 0), we can simplify the equation to x = (1/2)axt^2. Plugging in the values, we get 15 = (1/2)(2.0)t^2, which simplifies to t^2 = 15/2, or t ≈ 2.74 s.

Now that we know the time it takes for the particle to reach the x-coordinate of 15 m, we can find its y-velocity at that instant. The y-velocity can be given by the equation vy = v0y + ayt, where v0y is the initial velocity in the y-direction and ay is the acceleration in the y-direction. Since the particle has an initial y-velocity of 9.0 m/s in the positive y-direction (v0y = 9.0 m/s) and a constant y-acceleration of -4.0 m/s^2, we can plug in the values to get vy = 9.0 - (4.0)(2.74) ≈ -1.96 m/s.

Finally, we can find the speed of the particle at that instant by using the Pythagorean theorem. The speed can be given by the equation speed = sqrt(vx^2 + vy^2), where vx is the x-component of the velocity and vy is the y-component of the velocity. Plugging in the values, we get speed = sqrt((2.0)^2 + (-1.96)^2) ≈ 2.80 m/s.

So, the speed of the particle at the instant the x-coordinate is 15 m is approximately 2.80 m/s. Happy travels, little particle!

To find the speed of the particle at the instant when the x-coordinate is 15 m, we need to use the kinematic equations.

The equation we'll use is: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

Given:
Initial velocity, u = 9.0 m/s in the positive y-direction
Acceleration, a = (2.0i - 4.0j) m/s^2
Displacement, s = 15 m in the x-direction

Since the acceleration is only in the y-direction, it does not affect the speed in the x-direction. Hence, we can consider only the x-component of the velocity.

Since the acceleration is constant, the x-component of the velocity can be calculated using the equation:
v_x = u_x + at, where
v_x is the final velocity in the x-direction,
u_x is the initial velocity in the x-direction (which is 0),
a is the acceleration in the x-direction (which is 0), and
t is the time taken.

Since acceleration in the x-direction is 0, the x-component of the velocity remains constant with time.

So, v_x = u_x + at = 0 + 0 = 0 m/s.

Therefore, the x-component of the velocity is 0 m/s.

The speed of the particle is the magnitude of the total velocity, which is given by:
speed = |v| = sqrt(v_x^2 + v_y^2).

In this case, since v_x = 0 m/s, the speed is equal to the magnitude of the y-component of the velocity, v_y.

Given that u_y = 9.0 m/s, and the y-component of the acceleration j = -4.0 m/s^2, we can find the time it takes for the particle to reach an x-coordinate of 15 m.

We can use the equation of motion for displacement in the y-direction: s_y = u_y * t + 1/2 * a_y * t^2,
where s_y is the displacement in the y-direction,
u_y is the initial velocity in the y-direction,
a_y is the acceleration in the y-direction, and
t is the time of motion.

Since the particle starts at the origin, the displacement in the y-direction is zero.

Hence, 0 = 9 * t + 0.5 * (-4) * t^2.

Simplifying the equation, we get:
-2t^2 + 9t = 0.

Factoring out t, we get: t(9 - 2t) = 0.

So, either t = 0 or t = 9 / 2.

Since the particle is moving, the time taken cannot be zero, we discard t = 0 as the solution.

Therefore, t = 9 / 2 = 4.5 s.

Now, we can find the y-component of the final velocity using the equation: v_y = u_y + a_y * t,
where v_y is the final velocity in the y-direction.

Substituting the given values, we get:
v_y = 9.0 + (-4.0) * 4.5 = 9.0 - 18.0 = -9.0 m/s.

The y-component of the velocity is negative because the acceleration is in the negative y-direction.

Now, we can find the speed of the particle:
speed = |v| = sqrt(v_x^2 + v_y^2).

Substituting the values, we get:
speed = sqrt(0^2 + (-9.0)^2) = sqrt(0 + 81.0) = sqrt(81.0) = 9.0 m/s.

Therefore, the speed of the particle when the x-coordinate is 15 m is 9.0 m/s.

To find the speed of the particle at the instant when the x-coordinate is 15 m, we need to break down the velocity and acceleration vectors into their x and y components.

Given:
Initial velocity (v0) = 9.0 m/s in the positive y direction
Acceleration (a) = (2.0i - 4.0j) m/s^2
x-coordinate (x) = 15 m

Since the acceleration is constant, we can use the following equations of motion:

1. x = v0x * t + (1/2) * ax * t^2
2. v = v0 + at

Let's calculate the x and y components of the velocity vector first:

v0x = 0 m/s (since the initial velocity is only in the y direction)
v0y = 9.0 m/s

Now, let's calculate the time (t) at which the x-coordinate is 15 m:

x = v0x * t + (1/2) * ax * t^2
15 = 0 * t + (1/2) * 2.0 * t^2
15 = t^2
t = sqrt(15)

Using equation 2, we can find the y-component of the velocity at this time:

v = v0 + at
v = 9 + (-4 * sqrt(15))
v = 9 - 4 * sqrt(15)

Finally, to find the speed of the particle, we calculate the magnitude of the velocity vector:

speed = sqrt(vx^2 + vy^2)
speed = sqrt(0^2 + (9 - 4 * sqrt(15))^2)

Vf=Vi+at

Vf=9j+(2i-4j)t

displacement= origposition+at
now, in the i direction
15=O+2t or t=7.5 when x position is 15

put that into the first equation, solve for Vf.

speed= sqrt (xcomponent^2+ycomponent^2)

you get the x component, y component in the Vf equation