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Chapter 2: Problem 20

An object drops from a cliff that is 150 m high. The distance, \(d\), in metres, that the object has dropped at \(t\) seconds in modelled by \(d(t)=4.9 t^{2}\). a. Find the average rate of change of distance with respect to time from 2 s to 5 s. b. Find the instantaneous rate of change of distance with respect to time at \(4 \mathrm{s}\). c. Find the rate at which the object hits the ground to the nearest tenth.

Short Answer

Expert verified
a. 34.3 m/s b. 39.2 m/s c. 54.2 m/s

Step by step solution

01

Calculate Distance at t = 2s and t = 5s

To find the average rate of change, first calculate the distance the object has dropped at \(t = 2\) seconds and \(t = 5\) seconds. Using the formula \(d(t) = 4.9t^2\):\[ d(2) = 4.9 \times 2^2 = 19.6 \text{ m}\]\[ d(5) = 4.9 \times 5^2 = 122.5 \text{ m}\]
02

Compute Average Rate of Change

The average rate of change is computed using the formula for average rate of change, \( \frac{d(b) - d(a)}{b-a} \), where \(a\) and \(b\) are the initial and final times:\[ \text{Average rate of change} = \frac{d(5) - d(2)}{5 - 2} = \frac{122.5 - 19.6}{3} = \frac{102.9}{3} = 34.3 \text{ m/s} \]
03

Calculate Instantaneous Rate of Change at t = 4s

The instantaneous rate of change can be found by taking the derivative of \(d(t) = 4.9t^2\), which is \(\frac{d}{dt}d(t) = 9.8t\). Evaluate this derivative at \(t=4\) seconds:\[ \frac{d}{dt}d(4) = 9.8 \times 4 = 39.2 \text{ m/s} \]
04

Determine Final Time (When Object Hits the Ground)

To find when the object hits the ground, solve \(d(t) = 150\), since the cliff is 150 m high. Set up the equation:\[ 4.9t^2 = 150 \]Solving for \(t\):\[ t^2 = \frac{150}{4.9} = 30.6122 \]\[ t = \sqrt{30.6122} \approx 5.53 \text{ seconds}\]
05

Compute Rate of Change at t = 5.53s

Use the derivative to find the rate at which the object hits the ground by evaluating \( \frac{d}{dt}d(t) = 9.8t\) at \(t = 5.53\):\[ \frac{d}{dt}d(5.53) = 9.8 \times 5.53 \approx 54.2 \text{ m/s} \]

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Key Concepts

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

Understanding Average Rate of Change
The average rate of change is like a speedometer reading between two points in time. It's the change in position divided by the change in time.
It tells us how fast something was going over a period.
  • Calculate the position at two different times.
  • Subtract one position from the other to find the total change in position.
  • Divide by the time elapsed between these two points to find the average speed.
In our exercise, we found the object dropped 19.6 meters in 2 seconds and 122.5 meters in 5 seconds. Then, using \[ \frac{d(5) - d(2)}{5-2} = 34.3 \text{ m/s} \]we found the object averaged 34.3 m/s from 2 to 5 seconds.
Grasping Instantaneous Rate of Change
The instantaneous rate of change tells us how fast something is moving at an exact moment in time, much like the speedometer reading at that instant.
Mathematicians use derivatives to determine this rate.
  • This involves differentiating the function representing distance with respect to time.
  • Evaluate this derivative at the specific time you're interested in.
In the problem, we differentiate the function \(d(t) = 4.9t^2\), giving us \(\frac{d}{dt}d(t) = 9.8t\). Evaluating this at \(t=4\) seconds, we found the speed is \(39.2 \text{ m/s}\). This tells us at 4 seconds, the object is moving at 39.2 meters per second.
The Role of Derivatives
Derivatives play a crucial role in calculus, providing a formula for the slope at any given point on a curve. Think of it as the tool that gives us the instantaneous rate of change.
Calculus uses this tool to understand changes, especially in motion.
  • Derivatives can be seen as a limit of the average rate of change over smaller and smaller intervals.
  • The process of finding a derivative is called differentiation.
The derivative of \(d(t) = 4.9t^2\) is \(9.8t\). This translates motions equations into dynamic snapshots of speed at any given moment, as seen in the steps to solve the problem with derivatives.
Exploring Motion Equations
Motion equations help us describe how an object moves in space over time.
They share vital details about the object's position, velocity, and acceleration.
  • The function \(d(t) = 4.9t^2\) models how far the object has fallen with respect to time.
  • Velocity, or speed, is found using the derivative of the distance equation.
This allows us to compute the velocity - effectively Illuminating the object鈥檚 behavior from the cliff top to ground impact. Understanding each facet of motion enhances our grasp of what happens during the object鈥檚 fall.
Enhancing Problem-Solving Steps
Problem-solving in calculus involves breaking down the exercise into manageable pieces.
This structured approach makes complex problems more approachable and understandable.
  • Identify what is being asked - both rates, average and instantaneous, and when the object hits the ground.
  • Compute required values step-by-step, using known equations and differentiating where needed.
  • Verify and organize your calculations for clarity.
In our exercise, understanding each formula, calculation, and result formed the basis of solving. Mastering these steps means being equipped to tackle similar problems effectively, encouraging deeper comprehension of underlying principles.

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