Math Problem Statement

A model airplane has a position function given by i (t) = (-343 + 10t2 + 100t) + (-001t* - 4.912 + 25t)3. What is the airplane's displacement (in meters) from t = 1 s to t = 4 s?

Solution

The given position function of the model airplane is expressed as:

i(t)=(343+10t2+100t)i^+(0.001t34.912+25t)j^\mathbf{i}(t) = (-343 + 10t^2 + 100t)\hat{i} + (-0.001t^3 - 4.912 + 25t)\hat{j}

This is a vector function describing the position of the airplane at any time tt, where the i^\hat{i} and j^\hat{j} components represent the x and y coordinates, respectively.

To find the airplane's displacement from t=1st = 1 \, \text{s} to t=4st = 4 \, \text{s}, we will follow these steps:

Step 1: Find the position at t=1st = 1 \, \text{s}

Substitute t=1t = 1 into both the i^\hat{i}-component and the j^\hat{j}-component of the position function.

For i^\hat{i}-component: i(1)i^=343+10(1)2+100(1)=343+10+100=233\mathbf{i}(1)_{\hat{i}} = -343 + 10(1)^2 + 100(1) = -343 + 10 + 100 = -233

For j^\hat{j}-component: i(1)j^=0.001(1)34.912+25(1)=0.0014.912+25=20.087\mathbf{i}(1)_{\hat{j}} = -0.001(1)^3 - 4.912 + 25(1) = -0.001 - 4.912 + 25 = 20.087

Thus, the position at t=1st = 1 \, \text{s} is: i(1)=(233,20.087)\mathbf{i}(1) = (-233, 20.087)

Step 2: Find the position at t=4st = 4 \, \text{s}

Substitute t=4t = 4 into both the i^\hat{i}-component and the j^\hat{j}-component of the position function.

For i^\hat{i}-component: i(4)i^=343+10(4)2+100(4)=343+10(16)+400=343+160+400=217\mathbf{i}(4)_{\hat{i}} = -343 + 10(4)^2 + 100(4) = -343 + 10(16) + 400 = -343 + 160 + 400 = 217

For j^\hat{j}-component: i(4)j^=0.001(4)34.912+25(4)=0.001(64)4.912+100=0.0644.912+100=95.024\mathbf{i}(4)_{\hat{j}} = -0.001(4)^3 - 4.912 + 25(4) = -0.001(64) - 4.912 + 100 = -0.064 - 4.912 + 100 = 95.024

Thus, the position at t=4st = 4 \, \text{s} is: i(4)=(217,95.024)\mathbf{i}(4) = (217, 95.024)

Step 3: Calculate the displacement

The displacement is the difference between the positions at t=4st = 4 \, \text{s} and t=1st = 1 \, \text{s}.

For the i^\hat{i}-component: Δx=217(233)=217+233=450\Delta x = 217 - (-233) = 217 + 233 = 450

For the j^\hat{j}-component: Δy=95.02420.087=74.937\Delta y = 95.024 - 20.087 = 74.937

Thus, the displacement vector is: Δi=(450,74.937)\Delta \mathbf{i} = (450, 74.937)

Step 4: Calculate the magnitude of the displacement

The magnitude of the displacement vector is given by the formula: Δi=(Δx)2+(Δy)2|\Delta \mathbf{i}| = \sqrt{(\Delta x)^2 + (\Delta y)^2} Substituting the values: Δi=(450)2+(74.937)2=202500+5615.31=208115.31456.13m|\Delta \mathbf{i}| = \sqrt{(450)^2 + (74.937)^2} = \sqrt{202500 + 5615.31} = \sqrt{208115.31} \approx 456.13 \, \text{m}

Thus, the displacement of the airplane from t=1st = 1 \, \text{s} to t=4st = 4 \, \text{s} is approximately 456.13m456.13 \, \text{m}.


Would you like more details on any step?

Related Questions:

  1. How do you compute velocity from a position function?
  2. What is the significance of the direction of the displacement vector?
  3. How do you differentiate between displacement and distance?
  4. How would acceleration be calculated from this position function?
  5. What role do the components i^\hat{i} and j^\hat{j} play in vector functions?

Tip:

Always check if the components of your vector function involve time-dependent variables to avoid computational mistakes when finding positions or displacement over time.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Displacement
Position Function

Formulas

Displacement = Final Position - Initial Position
Magnitude of Displacement = sqrt((Δx)^2 + (Δy)^2)

Theorems

Pythagorean Theorem (for magnitude of displacement)

Suitable Grade Level

Grade 11-12 / Early College (Physics/Calculus)