Problem 17
Question
Velocity The height \(s(\text { in feet) at time } t\) (in seconds) of a silver dollar dropped from the top of the Washington Monument is given by $$ s=-16 t^{2}+555 $$ (a) Find the average velocity on the interval \([2,3] .\) (b) Find the instantaneous velocities when \(t=2\) and when \(t=3 .\) (c) How long will it take the dollar to hit the ground? (d) Find the velocity of the dollar when it hits the ground.
Step-by-Step Solution
Verified Answer
(a) The average velocity on the interval [2,3] is -80 feet/second. (b) The instantaneous velocities when t=2 and t=3 are -64 and -96 feet/second respectively. (c) It will take about 5.9 seconds for the dollar to hit the ground. (d) The velocity of the dollar when it hits the ground is approximately -189 feet/second.
1Step 1: Find the average velocity on the interval [2,3]
The formula for average velocity is \(\frac{\Delta s}{\Delta t}\). In other words, it's the change in position (height in this case) divided by the change in time. To calculate this, evaluate the function at both ends of the interval and subtract: \(\frac{s(3)-s(2)}{3-2} = \frac{-16*3^2+555 - (-16*2^2+555)}{1}\)
2Step 2: Find the instantaneous velocities when t=2 and when t=3
Instantaneous velocity corresponds to the derivative of the height over time function - in other words, it represents the rate of change of the position at a particular moment. To get the function representing the velocity, differentiate the given function: v(t) = s'(t) = -32t. To find the velocities at specific points in time (t=2 and t=3), substitute these values into the velocity function: v(2) = -32*2 and v(3) = -32*3
3Step 3: Find the time it takes for the dollar to hit the ground
The dollar hits the ground when the height is zero. To find this time, set the height function equal to 0 and solve for t: 0 = -16*t^2 + 555. Use algebra to isolate and solve for the variable t.
4Step 4: Find the velocity of the dollar when it hits the ground
This step requires finding the velocity at the time determined in Step 3. Substitute that time into the velocity function v(t) = -32t. This will give the velocity of the dollar when it hits the ground.
Key Concepts
Average Velocity FormulaInstantaneous VelocityPosition-Time Function DifferentiationVelocity at Ground Impact
Average Velocity Formula
The concept of average velocity is vital when studying motion in physics and calculus. It is a way to describe how quickly an object is changing its position over a period of time. To calculate average velocity, we use the simple formula: \[ v_{\text{avg}} = \frac{\Delta s}{\Delta t} \]Here, \( v_{\text{avg}} \) represents the average velocity, \( \Delta s \) is the change in position (also known as displacement), and \( \Delta t \) denotes the change in time. In the context of our exercise, where a silver dollar is dropped from the Washington Monument, we can find the average velocity between two time points by evaluating the position function: \[ s = -16t^2 + 555 \]at those points and then applying our average velocity formula. For example, between \( t = 2 \) seconds and \( t = 3 \) seconds, we calculate the average velocity by finding the difference in height at these times and dividing by the time interval, which is 1 second in this case. This will let us understand the 'overall' speed of the dollar over that one second span, ignoring the fluctuations that may happen within that interval.
Instantaneous Velocity
Unlike average velocity, which looks at an interval of time, instantaneous velocity captures how fast an object is moving at a specific moment. It is effectively the speed and direction of an object at a particular instant. To find this in calculus, we take the derivative of the position-time function, which gives us the velocity-time function. From the original problem, the height function given is \[ s = -16t^2 + 555. \]The derivative of this function with respect to time, denoted as \( s'(t) \) or \( v(t) \), is \[ v(t) = -32t. \]This function tells us the instantaneous velocity of the silver dollar for any value of \( t \). By plugging in the values of \( t \) at the moments of interest, such as \( t=2 \) and \( t=3 \), we can calculate how fast the dollar was moving at those specific seconds. It's this precise approach that differentiates instantaneous velocity from the broader perspective of average velocity.
Position-Time Function Differentiation
The act of deriving the instantaneous velocity from a position-time function represents one of the core applications of differentiation in calculus. Differentiation allows us to find the rate at which one variable changes with respect to another. In the case of our silver dollar, the position-time function is a downward parabola, expressed as \[ s = -16t^2 + 555. \]Differentiating this function with respect to time gives us \[ v(t) = s'(t) = \frac{d}{dt}(-16t^2 + 555) = -32t. \]This newly derived function expresses velocity as a function of time and reveals the instantaneous rate of change of the dollar’s height, that is, its velocity at any given moment during its fall. The power of differentiation lies in its ability to transform a static position equation into a dynamic velocity equation, offering a clear and precise look into the object's motion at each infinitesimal point in time.
Velocity at Ground Impact
Determining the velocity at ground impact is a compelling question in the study of motion, as it represents the speed just before an object collides with the Earth. Calculating this requires knowing two pieces of information: the time when the object hits the ground and the instantaneous velocity function. From our earlier steps, we've determined that we can set the height function equal to zero to find out when the object lands:\[ 0 = -16t^2 + 555. \]Through algebraic manipulation, one can solve for the time \( t \) at which the dollar reaches ground level. Then, by substituting this value of \( t \) back into our velocity-time function, \[ v(t) = -32t, \]we can find out the velocity at that precise moment. This gives us a numerical representation of the speed at which the silver dollar strikes the ground, a crucial piece of information for understanding the kinematics of objects under the influence of gravity.
Other exercises in this chapter
Problem 16
Use the limit definition to find the slope of the tangent line to the graph of \(f\) at the given point. $$ f(x)=2 x+4 ;(1,6) $$
View solution Problem 16
Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$\begin{array}{ll}{\text
View solution Problem 17
Find the derivative of the function. $$ y=4 t^{4 / 3} $$
View solution Problem 17
Moving Ladder A 25 -foot ladder is leaning against a house (see figure). The base of the ladder is pulled away from the house at a rate of 2 feet per second. Ho
View solution