Math Problem Statement

35. Average velocity The velocity in m/s of an object moving along a line over the time interval [0, 6] is v(t) = t^2 + 3t. Find the average velocity of the object over this time interval. 36. Average velocity A rock is launched vertically upward from the ground with a speed of 64 ft/s. The height of the rock (in ft) above the ground after t seconds is given by the function s(t) = -16t^2 + 64. Find its average velocity during its flight.

Solution

Let's break down the two problems you have shared:

35. Average velocity

The velocity v(t)v(t) is given as v(t)=t2+3tv(t) = t^2 + 3t over the time interval [0,6][0, 6].

To find the average velocity:

The average velocity is given by the formula:

Average velocity=1baabv(t)dt\text{Average velocity} = \frac{1}{b-a} \int_{a}^{b} v(t) \, dt

where a=0a = 0, b=6b = 6, and v(t)=t2+3tv(t) = t^2 + 3t.

So, the formula becomes:

Average velocity=16006(t2+3t)dt\text{Average velocity} = \frac{1}{6 - 0} \int_0^6 (t^2 + 3t) \, dt

Let me calculate this for you.

36. Average velocity of a rock

The position s(t)s(t) is given by the equation s(t)=16t2+64s(t) = -16t^2 + 64. You are asked to find the average velocity of the rock during its flight.

To find the average velocity:

We use the formula:

Average velocity=s(t2)s(t1)t2t1\text{Average velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}

where s(t)=16t2+64s(t) = -16t^2 + 64, t1t_1 is the time at launch, and t2t_2 is the time when the rock hits the ground.

  1. At t1=0t_1 = 0, s(0)=64s(0) = 64 feet (the initial height).
  2. To find t2t_2, set s(t)=0s(t) = 0 to determine when the rock hits the ground:

16t2+64=0-16t^2 + 64 = 0

Let me solve both these steps for you.Here are the solutions for the problems:

Problem 35:

The average velocity of the object over the time interval [0,6][0, 6] is: Average velocity=21m/s\text{Average velocity} = 21 \, \text{m/s}

Problem 36:

The average velocity of the rock during its flight (from launch to when it hits the ground) is: Average velocity=32ft/s\text{Average velocity} = -32 \, \text{ft/s} (The negative sign indicates that the average motion is downward.)

Would you like more details on any of these steps or have further questions?


Here are 5 related questions to expand your understanding:

  1. How do you interpret the physical meaning of average velocity when it is negative?
  2. How would the process change if you were given a non-quadratic velocity function?
  3. Can you find the total distance traveled by the object in problem 35 using calculus?
  4. What happens if we extend the time interval beyond 6 seconds for problem 35?
  5. How would the velocity function change for the rock in problem 36 if air resistance was considered?

Tip: Always ensure to correctly identify the time intervals when calculating average velocity. The interpretation of the results depends heavily on the limits chosen!

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Math Problem Analysis

Mathematical Concepts

Calculus
Kinematics
Average Velocity

Formulas

Average velocity = (1 / (b - a)) * ∫[a to b] v(t) dt
s(t) = -16t^2 + 64
Average velocity = (s(t2) - s(t1)) / (t2 - t1)

Theorems

Fundamental Theorem of Calculus
Kinematic Equations

Suitable Grade Level

Grades 11-12