At time t = 0s, a puck is sliding on a horizontal table with a velocity 3.00 m/s, 65.0 above the +x axis. As the puck slides, a constant acceleration acts on it that was the following components: ax = 0.460m/s2 and ay = 0.980m/s2. what is the velocity of the puck at time t = 1.50s?
If the x and y axes are on the horizontal table, how can the puck be "above" the x axis and sliding at the same time?
Perhaps the 65.0 refers to the direction angle in degrees, measured from the x axis.
In that case, at t = 1.50 s,
Vx = 3.00 cos 65 + 0.460 * 1.50
Vy = 3.00 sin 65 + 0.980 * 1.50
Compute both and then take the square root of the sum of the squares for the magnitude of the velocity (speed)
the 65 does not represent the degrees. . .that is why i am so confused on this problem
To find the velocity of the puck at time t = 1.50s, we can use the equations of motion.
The initial velocity of the puck is given as 3.00 m/s at an angle of 65.0 degrees above the +x axis. We can break down this initial velocity into its x-component and y-component.
Given:
Initial velocity, u = 3.00 m/s
Angle, θ = 65.0 degrees
To find the x-component of velocity (vx), we use the equation:
vx = u * cos(θ)
vx = 3.00 m/s * cos(65.0 degrees)
vx ≈ 3.00 m/s * 0.4226
vx ≈ 1.2678 m/s
To find the y-component of velocity (vy), we use the equation:
vy = u * sin(θ)
vy = 3.00 m/s * sin(65.0 degrees)
vy ≈ 3.00 m/s * 0.9063
vy ≈ 2.7188 m/s
Now, we can use the equations of motion to find the velocity at time t = 1.50s.
The x-component of velocity remains constant, as there is no acceleration in the x-direction.
vx' = vx = 1.2678 m/s
The y-component of velocity changes due to the constant acceleration in the y-direction.
Using the equation:
vy' = vy + a * t
vy' = vy + (ay * t)
vy' = 2.7188 m/s + (0.980 m/s² * 1.50 s)
vy' = 2.7188 m/s + 1.4700 m/s
vy' ≈ 4.1888 m/s
Finally, we can calculate the new velocity using the Pythagorean theorem:
v = sqrt(vx'² + vy'²)
v = sqrt((1.2678 m/s)² + (4.1888 m/s)²)
v ≈ sqrt(1.6060 m²/s² + 17.4881 m²/s²)
v ≈ sqrt(19.0941 m²/s²)
v ≈ 4.37 m/s
Therefore, the velocity of the puck at time t = 1.50s is approximately 4.37 m/s.
To find the velocity of the puck at time t = 1.50s, we can use the equations of motion. The equation for velocity can be split into its x and y components, since we have both horizontal (x) and vertical (y) accelerations.
First, let's find the x-component of the velocity. We know that the initial velocity in the x-direction (vx0) is 3.00 m/s, and the acceleration in the x-direction (ax) is 0.460 m/s^2. We can use the equation:
vx = vx0 + ax * t
Plugging in the values, we get:
vx = 3.00 m/s + (0.460 m/s^2) * 1.50s
= 3.00 m/s + 0.690 m/s
= 3.69 m/s
Now, let's find the y-component of the velocity. We know that the initial velocity in the y-direction (vy0) is 65.0 m/s, and the acceleration in the y-direction (ay) is 0.980 m/s^2. We can again use the equation:
vy = vy0 + ay * t
Plugging in the values, we get:
vy = 65.0 m/s + (0.980 m/s^2) * 1.50s
= 65.0 m/s + 1.470 m/s
= 66.47 m/s
Finally, we can find the magnitude of the velocity vector using the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)
Plugging in the values, we get:
v = sqrt((3.69 m/s)^2 + (66.47 m/s)^2)
= sqrt(13.6161 m^2/s^2 + 4416.7009 m^2/s^2)
= sqrt(4429.317 m^2/s^2)
≈ 66.6 m/s
Therefore, the velocity of the puck at t = 1.50s is approximately 66.6 m/s.