Math Problem Statement

The displacement (in centimeters) of a particle s moving back and forth along a straight line is given by the equation s = 5 sin(πœ‹t) + 2 cos(πœ‹t), where t is measured in seconds. (Round your answers to two decimal places.) (a) Find the average velocity during each time period.[1, 1.01]

[1, 1.001]When calculating the average velocity, you are actually calculating the slope of the secant line between the two given points. Use the displacement formula for each given time to calculate the difference in height for the given time period, and divide by the change in time.

Solution

To calculate the average velocity of the particle over the given time intervals, we use the formula for average velocity:

vavg=s(t2)βˆ’s(t1)t2βˆ’t1v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}

where:

  • s(t)s(t) is the displacement at time tt.
  • t1t_1 and t2t_2 are the initial and final times of the interval.

Displacement formula:

s(t)=5sin⁑(Ο€t)+2cos⁑(Ο€t)s(t) = 5 \sin(\pi t) + 2 \cos(\pi t)

(a) Find the average velocity for the interval [1, 1.01].

  1. Calculate s(1)s(1): s(1)=5sin⁑(Ο€β‹…1)+2cos⁑(Ο€β‹…1)=5sin⁑(Ο€)+2cos⁑(Ο€)s(1) = 5 \sin(\pi \cdot 1) + 2 \cos(\pi \cdot 1) = 5 \sin(\pi) + 2 \cos(\pi) Since sin⁑(Ο€)=0\sin(\pi) = 0 and cos⁑(Ο€)=βˆ’1\cos(\pi) = -1, we have: s(1)=5(0)+2(βˆ’1)=βˆ’2s(1) = 5(0) + 2(-1) = -2

  2. Calculate s(1.01)s(1.01): s(1.01)=5sin⁑(Ο€β‹…1.01)+2cos⁑(Ο€β‹…1.01)s(1.01) = 5 \sin(\pi \cdot 1.01) + 2 \cos(\pi \cdot 1.01) We can compute this using approximate values of the sine and cosine functions.

Let's compute the values for both steps.For the interval [1,1.01][1, 1.01]:

  • s(1)=βˆ’2.00 cms(1) = -2.00 \, \text{cm}
  • s(1.01)=βˆ’2.16 cms(1.01) = -2.16 \, \text{cm}

The average velocity is approximately:

vavg=βˆ’15.61 cm/sv_{\text{avg}} = -15.61 \, \text{cm/s}

(b) Now, let's compute the average velocity for the interval [1,1.001][1, 1.001]:

  1. We already have s(1)=βˆ’2s(1) = -2.

  2. Now, calculate s(1.001)s(1.001):

s(1.001)=5sin⁑(Ο€β‹…1.001)+2cos⁑(Ο€β‹…1.001)s(1.001) = 5 \sin(\pi \cdot 1.001) + 2 \cos(\pi \cdot 1.001)

I'll compute this next.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, you can manually calculate the displacement s(1.001)s(1.001) using the formula:

s(t)=5sin⁑(Ο€t)+2cos⁑(Ο€t)s(t) = 5 \sin(\pi t) + 2 \cos(\pi t)

Once you have that, you can use the formula for average velocity:

vavg=s(1.001)βˆ’s(1)1.001βˆ’1v_{\text{avg}} = \frac{s(1.001) - s(1)}{1.001 - 1}

Let me know if you'd like any further assistance or explanations! Would you like more help with these calculations?

Here are some related questions:

  1. What is the instantaneous velocity at t=1t = 1?
  2. How do we calculate the maximum speed of the particle?
  3. What happens to the average velocity as the time interval decreases?
  4. Can we find the time when the particle is at rest?
  5. How does changing the coefficients in the equation affect the motion?

Tip: The average velocity formula is similar to calculating the slope between two points on a graph!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Calculus
Average Velocity

Formulas

s(t) = 5 sin(Ο€t) + 2 cos(Ο€t)
Average velocity formula: v_avg = (s(t2) - s(t1)) / (t2 - t1)

Theorems

Sine and Cosine functions
Secant line slope

Suitable Grade Level

Grades 11-12 (Pre-Calculus/Calculus)