Problem 16 - The Calculus with Analytic Geometry

Question

A rocket is fired vertically upward, and it is \(s \mathrm{ft}\) above the ground \(t \mathrm{sec}\) after being fired, where \(s=560 t-16 t^{2}\) and the positive direction is upward. Find (a) the velocity of the rocket 2 sec after being fired, and (b) how long it takes for the rocket to reach its maximum height.

Step-by-Step Solution

Answer

The velocity 2 seconds after being fired is 496 ft/sec. It takes 17.5 seconds for the rocket to reach its maximum height.

Step 1: Find the velocity function

The velocity of the rocket is the first derivative of the position function with respect to time. The position function is given by \[s(t) = 560t - 16t^2\] Now, find the derivative of the position function to get the velocity function,\[v(t) = \frac{d}{dt}[560t - 16t^2] = 560 - 32t\]

Step 2: Calculate the velocity at 2 seconds

Substitute \(t = 2\) into the velocity function to find the velocity of the rocket 2 seconds after it is fired. \[v(2) = 560 - 32(2) = 560 - 64 = 496\] Therefore, the velocity of the rocket 2 seconds after being fired is 496 ft/sec.

Step 3: Find the time to reach maximum height

The rocket reaches its maximum height when the velocity is zero. Set the velocity function equal to zero and solve for \(t\). \[0 = 560 - 32t\] Add 32t to both sides: \[32t = 560\] Divide both sides by 32: \[t = \frac{560}{32} = 17.5\] Therefore, it takes 17.5 seconds for the rocket to reach its maximum height.

Key Concepts

Velocity Calculation

To find the velocity of the rocket at a specific time, we need to determine the rate of change of the position function with respect to time. This is done by finding the derivative of the position function. In our example, the position function is given by: \[s(t) = 560t - 16t^2\] The derivative of this function with respect to time gives us the velocity function, which is \[v(t) = 560 - 32t\]. We can now find the velocity at any given time by substituting the time value into the velocity function. For instance, to find the velocity 2 seconds after the rocket is fired, we substitute \(t = 2\) into the velocity function: \[v(2) = 560 - 32(2) = 496\]. Thus, the velocity of the rocket 2 seconds after being fired is 496 feet per second. This shows how the velocity calculation involves determining the rate at which the rocket's position changes over time.

Derivative of Position Function

The derivative of the position function gives us useful information about the velocity of the projectile. In our example, the function describing the rocket's position is: \[s(t) = 560t - 16t^2\]. To find the velocity, we take the derivative with respect to time \(t\). The general rules of differentiation are applied here: for a term such as \(At^n\), where \(A\) is a constant and \(n\) is a power, the derivative is given by \(nAt^{n-1}\). Applying this to our position function, we get: \[v(t) = \frac{d}{dt}[560t - 16t^2] = 560 - 32t\]. This function now represents the velocity of the rocket at any time \(t\). Understanding derivatives allows us to translate a position function into a velocity function, which is essential in analyzing motion in calculus.

Maximum Height Determination

The rocket reaches its maximum height when its velocity is zero. This means we need to find the time \(t\) at which the velocity function equals zero. Our velocity function is \[v(t) = 560 - 32t\]. Setting the velocity equal to zero, we solve for \(t\): \[0 = 560 - 32t\]. Add \(32t\) to both sides: \[32t = 560\]. Divide both sides by 32: \[t = \frac{560}{32} = 17.5\]. Therefore, the rocket reaches its maximum height 17.5 seconds after being fired. Understanding how to find the maximum height involves recognizing the importance of setting the velocity to zero and solving the resulting equation. This process helps identify critical moments in the motion, such as when the rocket moves from ascending to descending.