Question

A lunar lander is descending toward the moon's surface. Until the lander reaches the surface, its height above the surface of the moon is given by \(y(t) = b - ct + dt^2\) , where \(b =\) 800 m is the initial height of the lander above the surface, \(c =\) 60.0 m/s, and \(d =\) 1.05 m/s\(^2\). (a) What is the initial velocity of the lander, at \(t =\) 0? (b) What is the velocity of the lander just before it reaches the lunar surface?

Short answer

(a) The initial velocity is -60.0 m/s. (b) Just before impact, the velocity is approximately 62.51 m/s.

Step-by-step solution

Understanding Initial Velocity

To find the initial velocity of the lander, we look at the velocity function derived from the height function. The velocity is the first derivative of the height, given as:\[ v(t) = \frac{dy}{dt} = -c + 2dt \]Evaluating this at time \( t = 0 \), we find the initial velocity:\[ v(0) = -c = -60.0 \text{ m/s} \]

Finding Velocity at Surface

To find the velocity of the lander just before it reaches the lunar surface, set \( y(t) = 0 \) and solve for \( t \):\[ 0 = b - ct + dt^2 \]Substitute values into the equation:\[ 0 = 800 - 60t + 1.05t^2 \]This quadratic equation can be solved using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1.05 \), \( b = -60 \), and \( c = 800 \). Calculating the discriminant and solving:\[ t \approx 58.34 \text{ s} \] (we take only the positive root since time cannot be negative).

Calculate Velocity Just Before Impact

Using the found time \( t = 58.34 \text{ s} \), substitute back to the velocity equation:\[ v(t) = -c + 2dt \]\[ v(58.34) = -60 + 2 \times 1.05 \times 58.34 \]\[ v(58.34) \approx 62.51 \text{ m/s} \]

Key concepts

Initial Velocity Determination

Understanding the initial velocity of a moving object like a lunar lander is crucial for predicting its motion. The initial velocity represents the rate of change of position at the start of observation. In our case, it relates directly to how quickly the lander begins its descent when time equals zero.
For the lunar lander, the initial velocity can be found by differentiating the height function. The given height function is:
\[ y(t) = b - ct + dt^2 \]Here, the initial conditions are at 800m above the lunar surface with a given formula where coefficients define starting height, drag, and thrust effects.

• The initial velocity at time \( t = 0 \) is obtained by plugging \( t = 0 \) into the derivative of this function (the velocity function).
• Thus, the initial velocity formula simplifies to the coefficient \(-c\), leading to an initial velocity of \(-60.0\; \text{m/s}\).

This tells us that the lander begins its journey downwards (hence the negative sign) at a speed of 60.0 m/s.

Derivative Application in Physics

In physics, derivatives are a powerful tool used to understand how physical quantities change over time. Here, we apply the derivative to assess how the lander's height changes to understand its velocity. Let's break this down further.
The height function \( y(t) \) is differentiated to find the velocity function \( v(t) \):

• The derivative of \( b - ct + dt^2 \) with respect to time gives us \( -c + 2dt \).
• This means the lander's velocity is both constant due to \(-c\) and variable with \(2dt\).

Through this derivative, we can predict how the velocity evolves at every point in time.
Understanding derivatives in this context is vital as it offers insights into real-world applications like motion prediction, allowing us to effectively manage space missions by foreseeing trajectory outcomes.

Quadratic Equation Solving

Quadratic equations frequently appear in physics, often describing paths of objects under various forces. Our task is to find when the lander reaches the surface, which is a typical application of solving such an equation.
The equation \( 0 = b - ct + dt^2 \) arises from setting the height \( y(t) \) to zero:

• It's a quadratic form representing the lander’s path to the surface.
• Solving it requires our known quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where, in the context of our problem: \( a = 1.05 \), \( b = -60 \), and \( c = 800 \).

After computation, the realistic solution yields a time of approximately \( 58.34\; \text{s} \), indicating when the lander will touch lunar ground.
This computation not only enables precise event timing but also reflects how quadratic equations model real-world scenarios in space physics.

Velocity Calculation

Determining the velocity of the lunar lander just before it reaches the surface is essential for ensuring a safe landing. Thankfully, once we know when it reaches the surface, we can substitute back into the velocity formula.
After finding \( t = 58.34\; \text{s} \), we substitute into the previously derived velocity function:

• \( v(t) = -c + 2dt \)
• Calculating this gives: \(-60 + 2 \times 1.05 \times 58.34 \approx 62.51\; \text{m/s}\)

This is the speed at which the lander will meet the lunar surface, a crucial factor that engineers must consider when designing landing systems to avert crashes.
Thus, knowing how to calculate velocities like this ensures our lunar missions can be planned and executed safely, harnessing physics' predictive power effectively.