Math Problem Statement

A cheetah chases a gazelle and you have captured the chase on video. You have digitized the video to get the following raw position vs time data. What is the acceleration for frame 3?

                Time (s)      X position (m)

Frame 1 1 10.52

Frame 2 2.11 20.04

Frame 3 3.28 36.51

Frame 4 4.15 56.4

Frame 5 5.07 81.63

Solution

To calculate the acceleration for frame 3, we need to determine the acceleration between frames 2 and 4. Acceleration can be calculated as the change in velocity over time. Here's how we can break it down:

Step 1: Calculate velocity at frame 2, 3, and 4

The velocity at each frame can be approximated by the change in position divided by the change in time between consecutive frames:

Velocity=ΔXΔt\text{Velocity} = \frac{\Delta X}{\Delta t}

Where ΔX\Delta X is the change in position, and Δt\Delta t is the change in time.

  1. Velocity between Frame 2 and Frame 3:

    • ΔX=X3X2=36.5120.04=16.47m\Delta X = X_3 - X_2 = 36.51 - 20.04 = 16.47 \, \text{m}
    • Δt=t3t2=3.282.11=1.17s\Delta t = t_3 - t_2 = 3.28 - 2.11 = 1.17 \, \text{s}

    Velocity23=16.471.17=14.06m/s\text{Velocity}_{2 \rightarrow 3} = \frac{16.47}{1.17} = 14.06 \, \text{m/s}

  2. Velocity between Frame 3 and Frame 4:

    • ΔX=X4X3=56.436.51=19.89m\Delta X = X_4 - X_3 = 56.4 - 36.51 = 19.89 \, \text{m}
    • Δt=t4t3=4.153.28=0.87s\Delta t = t_4 - t_3 = 4.15 - 3.28 = 0.87 \, \text{s}

    Velocity34=19.890.87=22.86m/s\text{Velocity}_{3 \rightarrow 4} = \frac{19.89}{0.87} = 22.86 \, \text{m/s}

Step 2: Calculate the acceleration

Now that we have the velocities at frames 2, 3, and 4, we can calculate the acceleration by finding the change in velocity between frames 2 and 4, divided by the change in time between these frames:

Acceleration=ΔvΔt\text{Acceleration} = \frac{\Delta v}{\Delta t}

Where Δv\Delta v is the change in velocity, and Δt\Delta t is the change in time between frames 2 and 4.

  • Δv=Velocity34Velocity23=22.8614.06=8.8m/s\Delta v = \text{Velocity}_{3 \rightarrow 4} - \text{Velocity}_{2 \rightarrow 3} = 22.86 - 14.06 = 8.8 \, \text{m/s}
  • Δt=t4t2=4.152.11=2.04s\Delta t = t_4 - t_2 = 4.15 - 2.11 = 2.04 \, \text{s}

Acceleration=8.82.044.31m/s2\text{Acceleration} = \frac{8.8}{2.04} \approx 4.31 \, \text{m/s}^2

Conclusion:

The acceleration between frames 2 and 4 is approximately 4.31m/s24.31 \, \text{m/s}^2.

Let me know if you would like further explanation or if you have other questions!


Here are some follow-up questions to expand on the concept:

  1. How would you calculate the instantaneous acceleration if you had more data points?
  2. Why is it important to use velocity over small intervals when calculating acceleration?
  3. How does the motion of the cheetah change with respect to acceleration in this case?
  4. What would happen if you used a different method, like calculating average velocity over a larger interval?
  5. How does this acceleration compare to the typical acceleration of a cheetah in real-life chases?

Tip: When dealing with raw data like this, always ensure that you have a consistent time scale and units across the measurements to avoid errors in calculation.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Velocity
Acceleration

Formulas

Velocity = ΔX / Δt
Acceleration = Δv / Δt

Theorems

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Suitable Grade Level

Grades 9-11