How far from the origin is the puck after 5 seconds (m)?
Since I cant post internet sites,
The vx graph starts at the origin and has a slope of +8. The vy graph has a slope of 0 and starts at 30. The y axis is in cm/s and I am looking for total distance in meters. Thanks =]
To find the distance of the puck from the origin after 5 seconds, we need to consider both the x and y components of its velocity.
Let's break it down step by step:
1. Start by determining the x-component of the velocity (vx):
- Since the vx graph starts at the origin and has a slope of +8, we know that the x-component of the velocity is increasing by 8 cm/s every second.
- After 5 seconds, the change in x-component velocity is 8 cm/s * 5 s = 40 cm/s.
2. Next, let's calculate the y-component of the velocity (vy):
- The vy graph has a slope of 0, which means the y-component of the velocity remains constant at 30 cm/s.
3. Determine the total displacement of the puck in the x-direction (Δx):
- The displacement of an object can be calculated by multiplying its velocity by the time.
- Δx = vx * t = 40 cm/s * 5 s = 200 cm.
4. Convert the displacement from centimeters to meters:
- Since one meter is equal to 100 centimeters, you can divide the displacement by 100 to convert it to meters.
- Δx = 200 cm / 100 = 2 meters.
5. Calculate the total distance from the origin using the Pythagorean theorem:
- The total distance can be obtained by using the formula d = sqrt(Δx^2 + Δy^2), where Δx is the displacement in the x-direction and Δy is the displacement in the y-direction.
- Since the y-component of velocity is constant, the displacement in the y-direction is vy * t = 30 cm/s * 5 s = 150 cm.
- Δd = sqrt((2 m)^2 + (1.5 m)^2) = sqrt(4 m^2 + 2.25 m^2) = sqrt(6.25 m^2) = 2.5 meters.
Therefore, the puck is approximately 2.5 meters away from the origin after 5 seconds.