At time t = 0 s, a puck is sliding on a horizontal table with a velocity 3.60 m/s,
35.0° above the +x axis. As the puck slides, a constant acceleration acts on it
that has the following components: ax = –0.360 m/s2 and ay = –0.980 m/s2.
What is the velocity of the puck at time t = 1.50 s?
Well you're trying to find the v at 1.50. Which we'll call v2 brah. How does one get to v2 from the given though? Well you find the components of v1 first. v1 is 3.60 m/s as it is the first velocity we're dealing with. Hopefully you know how to find the components. You'll get 3.6sin35 and 3.6cos3.5 for v1y and v1x respectively. v1y is the y component of v1 and v1x is the x component. You'll get v1y as 2.06 and v1x as 2.95. Now use the kinematic equation v=vo+at. Only we're solving for v2's components now from v1's components. The equation becomes v2x=axt+v1x and v2y=ayt+v1y. Now we can use those accelerations they gave us. Set up your equations as follows. v2x=-.36(1.5)+2.95, and v2y=-.98(1.5)+2.06. You should get 2.41 and .59. Now just square both and root them. sqrt(2.41^2+.59^2), and you'll get v2. As doing this with any components of a vector will give you the vector's magnitude itself brah. You should get 2.48 as the result. To find the direction create a diagram for the components of v2 like you did v1. Instead of using 3.6 as the hypotenuse use 2.48, and set up your equation as cos(A)=v2x/v2, or sin(A)=v2y/v2, to get the angle. Plug in your prior numbers, cos(A)=2.41/2.48, and you'll get 13.6 when you do a cos inverse function on your calc.
Well, it seems like this puck is really going through some accelerated motion. I hope it's enjoying the ride!
To find the velocity of the puck at time t = 1.50 s, we can use some good old physics. Let's break this down step by step.
First, let's find the change in velocity (Δv) in both the x and y directions.
Δvx = ax * t,
where ax is the acceleration in the x-direction and t is the time given.
Plugging in the values, we have:
Δvx = (-0.360 m/s^2) * (1.50 s) = -0.54 m/s.
Similarly, for the y-direction:
Δvy = ay * t,
where ay is the acceleration in the y-direction and t is the time given.
Plugging in the values, we have:
Δvy = (-0.980 m/s^2) * (1.50 s) = -1.47 m/s.
Now, let's find the final velocity (v) of the puck. We can use the equation:
v = initial velocity (u) + Δv.
Given that the initial velocity is 3.60 m/s, let's find the final velocities in both the x and y directions.
vx = 3.60 m/s + (-0.54 m/s) = 3.06 m/s,
vy = 0 m/s + (-1.47 m/s) = -1.47 m/s.
To find the magnitude and direction of the final velocity, we can use the Pythagorean theorem and trigonometry.
The magnitude of the final velocity (v) is given by:
|v| = sqrt(vx^2 + vy^2).
Plugging in the values, we have:
|v| = sqrt((3.06 m/s)^2 + (-1.47 m/s)^2) = sqrt(9.3636 + 2.1609) = sqrt(11.5245) ≈ 3.39 m/s.
To find the direction of the final velocity (θ), we can use:
θ = arctan(vy / vx).
Plugging in the values, we have:
θ = arctan((-1.47 m/s) / (3.06 m/s)) ≈ -25.19°.
So, the velocity of the puck at time t = 1.50 s is approximately 3.39 m/s, at an angle of -25.19° relative to the positive x-axis.
Hope this helps you catch up with the speedy puck!
To find the velocity of the puck at time t = 1.50 s, we can use the kinematic equations. Let's break down the given information and solve step by step.
Given:
Initial velocity (u) = 3.60 m/s
Angle above the +x axis (θ) = 35.0°
Acceleration in the x-direction (ax) = -0.360 m/s^2
Acceleration in the y-direction (ay) = -0.980 m/s^2
Time (t) = 1.50 s
Step 1: Resolve the initial velocity into its x and y components.
The x-component of the initial velocity (ux) can be found using the formula:
ux = u * cos(θ)
Substituting the given values:
ux = 3.60 m/s * cos(35.0°)
Step 2: Calculate the change in x-position.
The change in x-position (Δx) can be found using the formula:
Δx = ux * t + (1/2) * ax * t^2
Substituting the given values:
Δx = (3.60 m/s * cos(35.0°)) * 1.50 s + (1/2) * (-0.360 m/s^2) * (1.50 s)^2
Step 3: Calculate the change in y-position.
The change in y-position (Δy) can be found using the formula:
Δy = uy * t + (1/2) * ay * t^2
Since there is no initial vertical velocity (uy = 0), the equation simplifies to:
Δy = (1/2) * ay * t^2
Substituting the given values:
Δy = (1/2) * (-0.980 m/s^2) * (1.50 s)^2
Step 4: Calculate the final velocity magnitude.
The final velocity magnitude (v) can be found using the Pythagorean theorem:
v = √((ux + Δx)^2 + (uy + Δy)^2)
Since uy = 0 and Δx and Δy have opposite signs due to the negative acceleration, the equation simplifies to:
v = √((ux + Δx)^2 + (Δy)^2)
Substituting the calculated values from step 2 and step 3:
v = √((3.60 m/s * cos(35.0°) + Δx)^2 + (Δy)^2)
Step 5: Calculate the final velocity direction.
The final velocity direction (θ') can be found using the inverse tangent:
θ' = atan(Δy / (ux + Δx))
Substituting the calculated values from step 2 and step 3:
θ' = atan(Δy / (3.60 m/s * cos(35.0°) + Δx))
The final answer is the magnitude (v) calculated in step 4 and the direction (θ') calculated in step 5.
To find the velocity of the puck at t = 1.50 s, we will use the equations of motion for constant acceleration.
The initial velocity of the puck v0 can be broken down into its x and y components using the given angle.
vx0 = v0 * cos(θ)
vy0 = v0 * sin(θ)
Given:
v0 = 3.60 m/s
θ = 35.0°
Using these values, we can find the initial x and y components of velocity.
vx0 = 3.60 m/s * cos(35.0°) = 2.944 m/s
vy0 = 3.60 m/s * sin(35.0°) = 2.044 m/s
Now, let's find the final velocity vx and vy at t = 1.50 s.
In the x-direction:
vx = vx0 + ax * t
Given:
ax = -0.360 m/s^2
t = 1.50 s
Substituting the values, we get:
vx = 2.944 m/s + (-0.360 m/s^2) * 1.50 s = 2.394 m/s
In the y-direction:
vy = vy0 + ay * t
Given:
ay = -0.980 m/s^2
t = 1.50 s
Substituting the values, we get:
vy = 2.044 m/s + (-0.980 m/s^2) * 1.50 s = 0.764 m/s
Finally, we can find the magnitude and direction of the final velocity using the Pythagorean theorem and trigonometry:
The magnitude of the final velocity is:
v = sqrt(vx^2 + vy^2)
Substituting the values, we get:
v = sqrt((2.394 m/s)^2 + (0.764 m/s)^2) = 2.501 m/s
The direction of the final velocity is given by:
θ = arctan(vy/vx)
Substituting the values, we get:
θ = arctan(0.764 m/s / 2.394 m/s) = 18.4°
Therefore, the velocity of the puck at t = 1.50 s is 2.501 m/s at 18.4° above the +x axis.