Problem 13
Question
If a ball is thrown into the air with a velocity of 40 \(\mathrm{ft} / \mathrm{s}\) , its height (in feet) after t seconds is given by $$\mathrm{y}=40 \mathrm{t}-16 \mathrm{t}^{2}$$ . Find the velocity when $$\mathrm{t}=2$ .$$
Step-by-Step Solution
Verified Answer
The velocity at \( t = 2 \) is \(-24 \) ft/s.
1Step 1: Understanding the formula
The formula given is the height of a ball thrown into the air: \( y = 40t - 16t^2 \). This represents the height, \( y \), as a function of time, \( t \). To find velocity at any time, we need the derivative of this height function.
2Step 2: Find the derivative for velocity
To find the velocity, take the derivative of the height function: \( y = 40t - 16t^2 \). The derivative with respect to \( t \) is \( v(t) = \frac{dy}{dt} = 40 - 32t \). This represents the velocity at any time \( t \).
3Step 3: Calculate the velocity at \( t = 2 \)
Now substitute \( t = 2 \) into the velocity equation \( v(t) = 40 - 32t \). So, \( v(2) = 40 - 32 \times 2 \).
4Step 4: Final calculation
By calculating the expression \( 40 - 32 \times 2 \), we get \( 40 - 64 = -24 \). Therefore, the velocity of the ball at \( t = 2 \) is \( -24 \) ft/s.
Key Concepts
Velocity CalculationHeight FunctionDerivative of a Function
Velocity Calculation
When learning about velocity calculation, it's important to first grasp the connection between velocity and a moving object's displacement over time. In this context, velocity is the rate of change of displacement, calculated as the derivative of the position function with respect to time.
In our given problem, we have a ball thrown into the air. The velocity of the ball isn’t constant as it changes at every moment due to gravity. The task is to find out how fast the ball is moving (the velocity) at a specific moment in time, precisely at 2 seconds.
We start by determining the height function, in which the height of the ball is given by the equation \( y = 40t - 16t^2 \). To calculate how quickly this height changes over time, we take the derivative of the height function, leading us to the velocity equation \( v(t) = 40 - 32t \). From this derivative, we can find the velocity by substituting the specific time into this velocity equation. This gives us the snapshot of the speed and direction of the ball's movement at that exact moment.
In our given problem, we have a ball thrown into the air. The velocity of the ball isn’t constant as it changes at every moment due to gravity. The task is to find out how fast the ball is moving (the velocity) at a specific moment in time, precisely at 2 seconds.
We start by determining the height function, in which the height of the ball is given by the equation \( y = 40t - 16t^2 \). To calculate how quickly this height changes over time, we take the derivative of the height function, leading us to the velocity equation \( v(t) = 40 - 32t \). From this derivative, we can find the velocity by substituting the specific time into this velocity equation. This gives us the snapshot of the speed and direction of the ball's movement at that exact moment.
Height Function
The height function is crucial in problems involving objects thrown upwards or projected into the air. It's a mathematical model that represents how high the object is at any given moment after it’s thrown.
For instance, in our problem, the height function is modeled by the equation \( y = 40t - 16t^2 \). Here, the variable \( t \) represents time in seconds, and \( y \) represents height in feet.
There are two terms in the height function:
For instance, in our problem, the height function is modeled by the equation \( y = 40t - 16t^2 \). Here, the variable \( t \) represents time in seconds, and \( y \) represents height in feet.
There are two terms in the height function:
- 40t: This is the initial velocity term. It explains how the starting velocity of the throw (40 feet per second upwards) contributes to the height of the ball.
- -16t^2: This reflects the effect of gravity, slowing the ball down as it rises and then speeding it up as it falls back down. The coefficient 16 comes from half the acceleration due to gravity in feet per second squared.
Derivative of a Function
Derivatives are fundamental in calculus as they help analyze the behavior of functions, especially in determining rates of change.
The derivative of a function provides us with a powerful tool to calculate how a quantity changes over time or any other variable. Think of it as the function's speedometer. In this exercise, the height function \( y(t) = 40t - 16t^2 \) models the ball's height. To find how quickly this height changes—i.e., the velocity—we need the derivative.
By applying basic differentiation rules, the derivative height function is \( \, \frac{dy}{dt} = 40 - 32t \). This new function lets us find the velocity of the ball at any time \( t \). It describes not just how high the ball is, but the rate at which that height is changing at every moment.
In summary, derivatives transform a complex question—how fast something changes—into a simple tool you can use for precise calculation.
The derivative of a function provides us with a powerful tool to calculate how a quantity changes over time or any other variable. Think of it as the function's speedometer. In this exercise, the height function \( y(t) = 40t - 16t^2 \) models the ball's height. To find how quickly this height changes—i.e., the velocity—we need the derivative.
By applying basic differentiation rules, the derivative height function is \( \, \frac{dy}{dt} = 40 - 32t \). This new function lets us find the velocity of the ball at any time \( t \). It describes not just how high the ball is, but the rate at which that height is changing at every moment.
In summary, derivatives transform a complex question—how fast something changes—into a simple tool you can use for precise calculation.
Other exercises in this chapter
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