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Chapter 2: Problem 25

The position of a diver executing a high dive from a 10 -m platform is described by the position function $$s(t)=-4.9 t^{2}+2 t+10 \quad t \geq 0$$ where \(t\) is measured in seconds and \(s\) in meters. a. When will the diver hit the water? b. How fast will the diver be traveling at that time? (Ignore the height of the diver and his outstretched arms.)

Short Answer

Expert verified
a. The diver will hit the water after approximately \(1.42\) seconds. b. The diver's speed when hitting the water is approximately \(12.016\) meters/second.

Step by step solution

01

a. Finding the time when the diver hits the water

Given the position function, \(s(t)=-4.9t^2 + 2t + 10\), we need to find the time \(t\) when the diver reaches the water, i.e., when \(s(t) = 0\): \[ 0 = -4.9t^2 + 2t + 10 \] This is a quadratic equation. To solve it, we can either factorize the equation or use the quadratic formula. In this case, factorization seems difficult, so we will use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = -4.9\), \(b = 2\), and \(c = 10\). Plugging the values in the formula, we get: \[ t = \frac{-2 \pm \sqrt{2^2 - 4(-4.9)(10)}}{2(-4.9)} \] Solving this equation, we get two possible values of \(t\), but since the diver's position is above the water at \(t = 0\), only the positive value of \(t\) is the physically meaningful solution. Calculating the positive value, we get: \[ t \approx 1.42 \text{ seconds} \] Thus, the diver will hit the water after approximately \(1.42\) seconds.
02

b. Finding the speed at which the diver hits the water

To find the speed of the diver when they hit the water, we need to first find their velocity function, which is the derivative of the position function with respect to time: \[ s'(t) = \frac{d}{dt}(-4.9t^2 + 2t + 10) \] Differentiating the position function with respect to time, we get the velocity function: \[ v(t) = s'(t) = -9.8t + 2 \] Now we have the velocity function, we can find the speed at which the diver hits the water by evaluating the velocity function at the time when the diver reaches the water (\(t \approx 1.42 \text{ seconds}\)): \[ v(1.42) = -9.8(1.42) + 2 \approx -12.016 \text{ meters/second} \] Notice that the velocity is negative, as the diver is moving downward. Since we're asked for the speed, we will consider only the magnitude of the velocity: \[ \text{Speed} \approx |v(1.42)| = 12.016 \text{ meters/second} \] So, the diver is traveling at approximately \(12.016\) meters/second when they hit the water.

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Key Concepts

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

Quadratic Equation
A quadratic equation is a second-degree polynomial equation in a single variable, with the form:\[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the unknown variable. Quadratic equations are fundamental in algebra and arise in various applications like motion and geometry. In this diver's problem, our position function is a quadratic equation, showcasing a parabola when graphed.
  • The term \( ax^2 \) determines the parabola's concavity (open upwards or downwards).
  • The term \( bx \) affects the direction and steepness.
  • \( c \) is the y-intercept, where the graph intersects the vertical axis.
To solve a quadratic equation for its roots (where the graph hits the x-axis), you can either factorize, complete the square, or commonly use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula provides two solutions, considering the \( \pm \) symbol, indicating two potential points where the function equals zero. In our diver's case, only the positive time solution physically makes sense as time cannot be negative.
Derivative Calculation
Derivatives are a key concept in calculus. They represent the rate at which a function's value changes as its input changes. In simpler terms, they are used to find how quickly something is changing at any given point. The derivative of a function at a particular point gives the slope of the tangent line to the function at that point. In the diver's problem, we are interested in finding out how the diver's position changes over time, which brings us to the concept of differentiation:
  • The derivative of \( ax^n \) with respect to \( x \) is \( a \cdot n \cdot x^{n-1} \).
  • For constant terms, the derivative is zero.
Differentiating the diver's position function \( s(t) = -4.9t^2 + 2t + 10 \) with respect to time \( t \) yields the velocity function:\[ s'(t) = v(t) = -9.8t + 2 \]This process of finding the derivative is vital for understanding how quickly the position function's output (position) changes concerning time, effectively giving us the diver's instantaneous velocity.
Velocity Function
The velocity function tells us how fast an object's position is changing over time. For our diver, this function is derived from the position function through differentiation. A velocity function presents the rate of change in an object's position as it moves.For the diver's jump, the velocity function is:\[ v(t) = -9.8t + 2 \]
  • The coefficient \(-9.8\) indicates the acceleration due to gravity affecting the diver's fall.
  • The positive constant \(2\) reflects the initial upward velocity as the diver pushes off the platform.
When analyzing the diver's movement, the negative velocity value at \( t = 1.42 \) seconds denotes the diver's downward direction as they approach the water. By plugging in this time into the velocity function, we determine the exact speed of the diver right before they contact the water surface.
Position Function
A position function describes an object's location as a function of time. It's crucial for understanding where an object started and how it moves over time. This function directly links an input variable (time) with an output (position), offering a complete picture of an object's movement.In our exercise, the position function of the diver is:\[ s(t) = -4.9t^2 + 2t + 10 \]
  • The quadratic form \(-4.9t^2\) captures the effects of gravity, causing the diver to accelerate downwards.
  • The linear term \(2t\) reflects the initial upward momentum as the diver jumps.
  • The constant \(10\) represents the height of the diving platform.
At \( t = 0 \), the diver is at the top of the dive, 10 meters above the ground. As time progresses, the parabolic nature of the function tracks the height of the diver until they reach \( s(t) = 0 \), indicating they've hit the water. This function encapsulates the diver's entire trajectory from the platform to the water.

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