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Chapter 3: Problem 16

Refer to the following. Suppose a ball is thrown straight upward with an initial velocity (that is, velocity at the time of release) of \(128 \mathrm{ft} / \mathrm{sec}\) and that the point at which the ball is released is considered to be at zero height. Then the height \(s(t)\) in feet of the ball at time \(t\) in seconds is given by \(s(t)=-16 t^{2}+128 t .\) Let \(v(t)\) be the instantaneous velocity at time \(t\). Find \(v(t)\) for the indicated values of \(t\). \(v(5)\)

Short Answer

Expert verified
The velocity at 5 seconds is -32 ft/sec.

Step by step solution

01

Write the General Formula for Velocity

The velocity function \(v(t)\) is the derivative of the position function \(s(t)\). Therefore, we need to find the derivative of \(s(t) = -16t^2 + 128t\). This will give us \(v(t)\).
02

Differentiate the Position Function

To find \(v(t)\), differentiate \(s(t) = -16t^2 + 128t\). Using the power rule for differentiation, we get: \[ v(t) = \frac{d}{dt}[-16t^2 + 128t] = -32t + 128. \]
03

Evaluate the Velocity at t=5

Substitute \(t=5\) into the velocity function \(v(t) = -32t + 128\).\[ v(5) = -32(5) + 128 = -160 + 128 = -32. \]
04

Conclude the Velocity at t=5

The velocity \(v(t)\) at \(t=5\) is \(-32\) ft/sec. This is the instantaneous velocity of the ball at 5 seconds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Derivatives
The concept of a derivative is pivotal in calculus, especially when analyzing the motion of objects. A derivative measures how a function changes as its input changes. In simpler terms, it gives us the rate at which one quantity, such as height or position, changes with respect to another, such as time.
For our problem, the position of the ball, described by the function \(s(t) = -16t^2 + 128t\), changes as time progresses. The derivative of this position function allows us to determine how quickly the height of the ball changes over time.
  • A derivative expresses the "instantaneous rate of change."
  • It can be thought of as the slope of the tangent line to the curve at any given point.
  • For straightforward functions, techniques like the power rule can simplify the differentiation process.
Linking Derivatives to Velocity
Velocity in physics is essentially the rate of change of position. By differentiating a position function with respect to time, we obtain a function that describes this rate of change: the velocity function.
For the ball's motion described in the problem, we derived the velocity function \(v(t)\) from the position function \(s(t)\). This function characterizes how fast the ball's position changes per unit of time.
  • Velocity is the "first derivative" of the position function.
  • The sign of the velocity indicates the direction of motion.
  • Positive velocity suggests upward motion, whereas negative implies downward.
Analyzing the Position Function
The position function \(s(t) = -16t^2 + 128t\) models the height of the ball over time. It's crafted using the physics of motion under gravity, where the coefficient \(-16\) is derived from gravitational acceleration in ft/s虏.
The function helps us visualize the trajectory:
  • The term \(-16t^2\) reflects the acceleration due to gravity.
  • The term \(128t\) represents the initial velocity of the ball.
  • Zero height at release is indicated by the constant term being zero.
This quadratic form means the trajectory is parabolic, with the ball first rising, slowing down, and then descending.
Differentiation Process Explained
Differentiation is the technique used to derive the velocity function from the position equation. The simplest rule often applied is the "power rule," which makes it straightforward to compute derivatives of polynomials.
For a function like \(s(t) = -16t^2 + 128t\), the power rule lets us easily find the derivative:
  • For \(-16t^2\), multiply the exponent by the coefficient: \(-32t^{1}\).
  • For \(128t\), the derivative is simply the coefficient, \(128\).
  • Combine these to obtain \(v(t) = -32t + 128\).
This shows us how velocity changes at any point in time.
Understanding Instantaneous Velocity
Instantaneous velocity is essentially the derivative value evaluated at a specific time. It tells us the exact speed and direction of the moving object at that moment.
For \(t = 5\), substituting into the velocity equation \(v(t) = -32t + 128\), we found that \(v(5) = -32\) ft/sec. This indicates:
  • The ball is moving downwards.
  • Its speed is 32 ft/sec at precisely 5 seconds.
This concept is vital as it extends the idea of average velocity over an interval to a precise moment in time.

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