Math Problem Statement

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 2 sin(πt) + 3 cos(πt), where t is measured in seconds. (a) Find the average velocity during each time period: (i) [1, 2], (ii) [1, 1.1], (iii) [1, 1.01], (iv) [1, 1.001]. (b) Estimate the instantaneous velocity of the particle when t = 1.

Solution

We are given the displacement equation s(t)=2sin(πt)+3cos(πt)s(t) = 2 \sin(\pi t) + 3 \cos(\pi t), where ss is the displacement in centimeters, and tt is the time in seconds.

(a) Average Velocity

The average velocity vavgv_{avg} over a time interval [t1,t2][t_1, t_2] is given by:

vavg=s(t2)s(t1)t2t1v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}

We will compute this for the four intervals:

(i) [1, 2]

vavg=s(2)s(1)21v_{avg} = \frac{s(2) - s(1)}{2 - 1}

(ii) [1, 1.1]

vavg=s(1.1)s(1)1.11v_{avg} = \frac{s(1.1) - s(1)}{1.1 - 1}

(iii) [1, 1.01]

vavg=s(1.01)s(1)1.011v_{avg} = \frac{s(1.01) - s(1)}{1.01 - 1}

(iv) [1, 1.001]

vavg=s(1.001)s(1)1.0011v_{avg} = \frac{s(1.001) - s(1)}{1.001 - 1}

(b) Instantaneous Velocity at t=1t = 1

The instantaneous velocity is the derivative of the displacement function s(t)s(t). First, we calculate the derivative s(t)s'(t):

s(t)=ddt(2sin(πt)+3cos(πt))s'(t) = \frac{d}{dt}\left(2 \sin(\pi t) + 3 \cos(\pi t)\right)

Using the chain rule: s(t)=2πcos(πt)3πsin(πt)s'(t) = 2 \pi \cos(\pi t) - 3 \pi \sin(\pi t)

To find the instantaneous velocity at t=1t = 1, we substitute t=1t = 1 into s(t)s'(t):

s(1)=2πcos(π×1)3πsin(π×1)s'(1) = 2 \pi \cos(\pi \times 1) - 3 \pi \sin(\pi \times 1)

Now, let's calculate the values. I will solve these step by step.### Results:

(a) Average velocity for each time period:

  1. For [1,2][1, 2]: vavg=6cm/sv_{avg} = 6 \, \text{cm/s}
  2. For [1,1.1][1, 1.1]: vavg4.71cm/sv_{avg} \approx -4.71 \, \text{cm/s}
  3. For [1,1.01][1, 1.01]: vavg6.13cm/sv_{avg} \approx -6.13 \, \text{cm/s}
  4. For [1,1.001][1, 1.001]: vavg6.27cm/sv_{avg} \approx -6.27 \, \text{cm/s}

(b) Instantaneous velocity at t=1t = 1:

The instantaneous velocity is approximately v(1)6.28cm/sv(1) \approx -6.28 \, \text{cm/s}.


Do you need further details or clarifications on any part?
Here are some related questions:

  1. How does the instantaneous velocity relate to the average velocity as the time interval shrinks?
  2. Can you derive the derivative of the displacement function step by step?
  3. What is the physical meaning of negative velocity in this case?
  4. How would the results change if we used different trigonometric functions in the displacement formula?
  5. What is the behavior of the velocity as tt \to \infty?

Tip: Instantaneous velocity can be thought of as the limit of average velocity over smaller and smaller time intervals.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Derivatives
Average Velocity
Instantaneous Velocity

Formulas

s(t) = 2 sin(πt) + 3 cos(πt)
Average velocity formula: v_avg = (s(t2) - s(t1)) / (t2 - t1)
Instantaneous velocity formula: v(t) = s'(t)

Theorems

Chain Rule for differentiation
Limits and continuity (for instantaneous velocity)

Suitable Grade Level

Undergraduate or Advanced High School (Grades 11-12)