Question

The height \(s\) in feet of a ball above the ground at \(t\) seconds is given by \(s=-16 t^{2}+40 t+100\). (a) What is its instantaneous velocity at \(t=2\) ? (b) When is its instantaneous velocity 0 ?

Short answer

(a) The instantaneous velocity at \(t=2\) is -24 ft/s. (b) The instantaneous velocity is 0 at \(t = \frac{5}{4}\) seconds.

Step-by-step solution

Understand the Formula for Instantaneous Velocity

Instantaneous velocity is found by taking the derivative of the position function with respect to time. For this problem, the height function is given by \(s(t) = -16t^2 + 40t + 100\). We need to find \(\frac{ds}{dt}\), the derivative of \(s\) with respect to \(t\).

Find the Derivative of the Height Function

The derivative \(\frac{ds}{dt}\) is taken by differentiating each term of the height function \(s(t)\). Using standard rules of differentiation:\(-16t^2\) becomes \(-32t\),\(40t\) becomes \(40\),and the constant \(100\) becomes \(0\) since its derivative is zero.Thus, \(\frac{ds}{dt} = -32t + 40\).

Substitute to Find Instantaneous Velocity at \(t=2\)

To find the instantaneous velocity at \(t=2\), substitute \(t = 2\) into the derived velocity function, \(v(t) = -32t + 40\):\[v(2) = -32(2) + 40 = -64 + 40 = -24\]Thus, the instantaneous velocity at \(t=2\) is \(-24\) feet per second.

Solve for When Instantaneous Velocity is Zero

Set the derivative equal to zero to find when the velocity is zero:\(-32t + 40 = 0\).Rearrange to solve for \(t\):\[-32t = -40\]\[t = \frac{-40}{-32} = \frac{40}{32} = \frac{5}{4}\].The instantaneous velocity is zero at \(t = \frac{5}{4}\) seconds.

Key concepts

Understanding Derivatives

Derivatives are fundamental tools in calculus, often used to determine how a function changes at any given point. To find the instantaneous velocity of an object, we need to take the derivative of its position function. This gives us the instantaneous rate of change, or velocity, at a specific moment in time. In simpler terms, while the position function tells us where the object is, the derivative of this function tells us how fast the object is moving and in which direction.

Think of derivatives as a way to "zoom in" on a curve to see its slope at a single point:

• The slope of the tangent line to the curve at any point gives us the instantaneous rate of change.
• This tangential slope is essentially the velocity when dealing with motion.

By differentiating the position function, we're able to establish how the height changes over time, allowing us to analyze specifics like instantaneous speed.

Position Function Fundamentals

The position function describes the location of an object in terms of a given variable, often time. For our exercise, the position function is described by the equation:

\[ s(t) = -16t^2 + 40t + 100 \]

This equation models the height, in feet, of a ball above the ground over time in seconds:

• The term \(-16t^2\) represents the effect of gravity, pulling the ball downward.
• The term \(40t\) represents the initial upward velocity, giving the ball its initial boost.
• The constant \(100\) represents the initial height from which the ball was thrown.

As time progresses, these elements combine to define the parabolic trajectory of the ball's flight, detailing how its position changes from moment to moment.

Applying Differentiation Rules

To find the instantaneous velocity, we differentiate the position function using basic differentiation rules. The process involves breaking down each term of the function and applying the derivative to each:

1. **Power Rule**: Used for terms with variables raised to a power, like \(t^2\). Differentiate \(-16t^2\) to get \(-32t\).2. **Constant Multiple Rule**: The derivative of a constant times a function is the constant times the derivative of the function. For \(40t\), differentiation results in \(40\).3. **Constant Rule**: The derivative of a constant is zero, as seen with \(100\).

Putting these together, the derivative of the position function \(s(t)\) becomes:
\[ \frac{ds}{dt} = -32t + 40 \]
This expression shows the velocity of the ball. By substituting specific values of \t\, we can determine the ball’s speed at that precise moment. For example, substituting \(t = 2\), we find the instantaneous velocity to be \(-24\) feet per second. Additionally, by setting the derivative to zero, we can determine when the ball momentarily stops, which occurs at \(t = \frac{5}{4}\) seconds.