Question

Comparing velocities A stone is thrown vertically into the air at an initial velocity of \(96 \mathrm{ft} / \mathrm{s} .\) On Mars the height (in feet) of the stone above the ground after \(t\) seconds is \(s=96 t-6 t^{2},\) and on Earth it is \(s=96 t-16 t^{2} .\) How much higher will the stone travel on Mars than on Earth?

Short answer

Answer: The stone will travel 240 feet higher on Mars than on Earth.

Step-by-step solution

Finding the velocity function

To find the velocity function for both Earth and Mars, we will take the derivative of the given position functions with respect to time (t). For Mars: \(v_m(t) = \frac{d(96t - 6t^2)}{dt} = 96 - 12t\) For Earth: \(v_e(t) = \frac{d(96t - 16t^2)}{dt} = 96 - 32t\)

Finding the time at which the stone reaches maximum height

To find the time at which the stone reaches the maximum height, we will set each velocity function equal to 0. For Mars: \(v_m(t) = 0 \Rightarrow 96 - 12t = 0 \Rightarrow t = \frac{96}{12} = 8\) seconds For Earth: \(v_e(t) = 0 \Rightarrow 96 - 32t = 0 \Rightarrow t = \frac{96}{32} = 3\) seconds

Finding the maximum height for both Earth and Mars

We will substitute the time values we've found in Step 2 into the height functions to find the maximum height. For Mars: \(s_m = 96(8) - 6(8)^2 = 768 - 384 = 384\) feet For Earth: \(s_e = 96(3) - 16(3)^2 = 288 - 144 = 144\) feet

Finding the difference in maximum heights

Now we will subtract the maximum height on Earth from the maximum height on Mars: Difference in height: \(384 - 144 = 240\) feet So the stone will travel 240 feet higher on Mars than on Earth.

Key concepts

Derivative of Position Functions

The derivative of a position function with respect to time, commonly represented as \( s'(t) \), is the velocity function, which provides vital information about an object's motion. In the context of comparing the velocities of a projectile on Mars versus Earth, derivatives help us understand how the velocity changes over time.

For example, the derivative of the position function \( s = 96t - 6t^2 \) for a stone thrown on Mars is calculated as \( v_m(t) = 96 - 12t \). This resultant function indicates that the stone's velocity decreases by 12 feet per second every second it's in the air, due to the acceleration due to gravity on Mars. Similarly, on Earth, the derivative of \( s = 96t - 16t^2 \) gives us \( v_e(t) = 96 - 32t \), which tells us the stone on Earth will slow down faster due to a greater acceleration due to gravity.

Understanding the derivative allows students to predict the stone's motion, including its velocity at any given moment and when it reaches its peak height, which occurs when the velocity becomes zero.

Maximum Height Projectile

To analyze the maximum height achieved by a projectile, we take advantage of the fact that at the highest point, the velocity of the projectile is zero. By setting the velocity function to zero, we can find the time at which the stone reaches its peak.

On Mars, setting \( v_m(t) = 0 \) and solving for \( t \) gives us the time when the stone reaches its maximum height. Mathematically, it is expressed as \( 96 - 12t = 0 \), yielding \( t = 8 \) seconds. Conversely, on Earth, using \( v_e(t) = 0 \), the equation \( 96 - 32t = 0 \) gives us \( t = 3 \) seconds.

Once the times are found, substituting these back into the original position functions \( s = 96t - 6t^2 \) for Mars and \( s = 96t - 16t^2 \) for Earth provides the maximum heights, which are vital for comparing projectile motion across different gravitational fields.

Velocity Function in Calculus

The velocity function in calculus is an essential tool for understanding motion. It is derived from the position function, which describes the location of an object over time. The velocity function represents the rate at which the object's position is changing and is defined as the derivative of the position function with respect to time.

In our example, the velocity functions for a stone thrown on Mars and Earth, denoted as \( v_m(t) \) and \( v_e(t) \), respectively, not only signify the stone's speed at any given moment but also allow for the determination of the maximum height reached. They are key in calculating the momentary speed of the stone and, by extension, when the stone momentarily stops at the peak of its trajectory before descending.

Understanding the concept of a velocity function in calculus empowers students to solve real-world problems involving motion, ranging from simple projectile movements to complex dynamics in physics.