Problem 38 - Precalculus: Mathematics for Calculus

Question

Velocity on the Moon If an arrow is shot upward on the moon with a velocity of \(58 \mathrm{m} / \mathrm{s},\) its height (in meters) after \(t\) seconds is given by \(H=58 t-0.83 t^{2}\) (a) Find the instantaneous velocity of the arrow after 1 second. (b) Find the instantaneous velocity of the arrow when \(t=a\) (c) At what time \(t\) will the arrow hit the moon? (d) With what velocity will the arrow hit the moon?

Step-by-Step Solution

Answer

(a) 56.34 m/s, (b) 58 - 1.66a, (c) t ≈ 69.88 s, (d) ≈ -58.98 m/s.

Step 1: Understand the Problem

We need to find the instantaneous velocity of the arrow at various times given its height function on the moon, \( H(t) = 58t - 0.83t^2 \). Additionally, we need to determine when the arrow hits the moon and the velocity at that moment.

Step 2: Find the Derivative for Instantaneous Velocity

The instantaneous velocity is the derivative of the height function with respect to time, \( t \). Differentiate \( H(t) \):\[H'(t) = \frac{d}{dt}(58t - 0.83t^2) = 58 - 2(0.83)t = 58 - 1.66t\]This gives us the velocity function, \( v(t) = 58 - 1.66t \).

Step 3: Calculate Instantaneous Velocity at t = 1 second

Substitute \( t = 1 \) into the velocity function:\[v(1) = 58 - 1.66(1) = 58 - 1.66 = 56.34 \, \text{m/s}\]The instantaneous velocity after 1 second is \( 56.34 \, \text{m/s} \).

Step 4: Determine Instantaneous Velocity Formula at t = a

For \( t = a \), use the velocity function:\[v(a) = 58 - 1.66a\]This formula gives the instantaneous velocity at any time \( t = a \).

Step 5: Solve for Time When Arrow Hits the Moon

The arrow hits the moon when its height \( H(t) = 0 \). Set the height function equal to zero and solve for \( t \):\[58t - 0.83t^2 = 0 \t(58 - 0.83t) = 0\]This gives \( t = 0 \) or \( 58 - 0.83t = 0 \). Solving for the non-zero time:\[0.83t = 58 \t = \frac{58}{0.83} \approx 69.88\]Thus, the arrow hits the moon at approximately \( 69.88 \) seconds.

Step 6: Calculate Velocity When Arrow Hits the Moon

Substitute \( t = 69.88 \) into the velocity function:\[v(69.88) = 58 - 1.66(69.88)\]Calculate the velocity:\[v(69.88) \approx 58 - 116.9808 \approx -58.9808 \, \text{m/s}\]The arrow hits the moon with a velocity of approximately \( -58.98 \, \text{m/s} \). The negative sign indicates the direction is downward.

Key Concepts

Differentiation

Differentiation is a method used in calculus to find rates of change. It helps us understand how a function behaves by determining its derivative.
If we have a function that describes a quantity changing over time, its derivative will give us the rate at which it changes. Essentially, differentiation breaks down a big picture into small chunks, allowing us to evaluate how that picture changes at any specific point.

For the arrow shot upwards, we have the height function:

• The function is given by: \( H(t) = 58t - 0.83t^2 \) where \( t \) is time.
• To find the instantaneous velocity, differentiate \( H(t) \) to get \( H'(t) \). The derivative, \( H'(t) = 58 - 1.66t \), represents the velocity of the arrow at any time \( t \).

This tells us how fast and in which direction the arrow is moving at any point in time. So, whenever you need to calculate an instantaneous rate of change, differentiation is your tool.

Arrow Motion on the Moon

The motion of an arrow when shot on the Moon is quite intriguing due to the Moon's weaker gravity compared to Earth. This example allows us to see physics in action via mathematical equations.

When an arrow is shot upwards:

• The initial velocity is given by \( 58 \, \text{m/s} \).
• The motion is influenced by gravity, but it's weaker on the Moon, represented by the term \( -0.83t^2 \) in the height function.

The gravity on the Moon affects how quickly the arrow decelerates and comes back down. Now, to find when the arrow returns to the Moon's surface:

• We set the height function equal to zero: \( H(t) = 0 \).
• Solve for \( t \) and find the non-zero solution through \( 58t - 0.83t^2 = 0 \) which results in \( t \approx 69.88 \) seconds.

This calculation predicts the "landing" time based on the given equation, showing the symmetry in projectile motion where time up equals time down, adjusted by lunar gravity.

Physics Applications in Precalculus

Physics and precalculus often intersect when studying motion, as seen with the arrow's trajectory on the Moon. Understanding this linkage helps in expanding both mathematical skills and scientific understanding.

When studying projectile motion:

• Height formulas like \( H(t) = 58t - 0.83t^2 \) link initial velocity and gravitational effects.
• Instantaneous velocity is crucial for understanding speed and direction changes, accessible through derivatives.

This exercise demonstrates how precalculus tools like differentiation empower you to solve real-world physics problems.

By using the velocity function \( v(t) = 58 - 1.66t \), we calculate dynamic states of motion:

• The velocity at liftoff, time of impact, and velocity at impact (here returning downwards).

Thus, equations are not just abstract notions, but practical instruments that explain how objects interact with forces, such as gravity, around them.