Question

technical math: exercises 3-1

1. on the earth’s moon, gravity is only one - sixth as strong as it is on earth, so an object on the moon will fall one - sixth the distance it would fall on earth in the same time. this means that the gravity distance function for a falling object on the moon is

$s = \frac{16}{6}t^2$ or $s = \frac{8}{3}t^2$,
where $t$ represents time since the object is released, in seconds, and $s$ is the distance fallen, in ft.

a. how far does an object on the moon fall in 3 seconds?
$s =$

b. how long does it take an object on the moon to fall 96 feet?

c. graph the moon’s gravity function on a properly scaled and labeled coordinate axis or on a graphing calculator for $t = 0$ to $t = 5$.

d. use the graph to estimate how long it takes an object on the moon to fall 35 feet.

e. write an equation to determine the required time in part d. solve the equation algebraically.

f. how does your answer in part e compare to your estimate in part d?

Explanation:

Part a

Step1: Identify the formula

We use the formula \( s=\frac{8}{3}t^{2} \) (simplified form of \( \frac{16}{6}t^{2} \)) for the distance fallen on the Moon.

Step2: Substitute \( t = 3 \)

Substitute \( t = 3 \) into the formula: \( s=\frac{8}{3}\times(3)^{2} \).
First, calculate \( (3)^{2}=9 \). Then, \( \frac{8}{3}\times9 = 8\times3=24 \).

Step1: Set up the equation

We know \( s = 96 \) and the formula \( s=\frac{8}{3}t^{2} \). So, \( 96=\frac{8}{3}t^{2} \).

Step2: Solve for \( t^{2} \)

Multiply both sides by \( \frac{3}{8} \): \( t^{2}=96\times\frac{3}{8} \).
Calculate \( 96\times\frac{3}{8}=12\times3 = 36 \).

Step3: Solve for \( t \)

Take the square root of both sides: \( t=\sqrt{36}=6 \) (we take the positive root since time cannot be negative).

Step1: Determine the function

The function is \( s=\frac{8}{3}t^{2} \), with \( t\in[0,5] \).

Step2: Find key points

• When \( t = 0 \), \( s=\frac{8}{3}\times0^{2}=0 \), so the point is \( (0,0) \).
• When \( t = 1 \), \( s=\frac{8}{3}\times1^{2}=\frac{8}{3}\approx2.67 \), point \( (1,\frac{8}{3}) \).
• When \( t = 2 \), \( s=\frac{8}{3}\times4=\frac{32}{3}\approx10.67 \), point \( (2,\frac{32}{3}) \).
• When \( t = 3 \), \( s = 24 \) (from part a), point \( (3,24) \).
• When \( t = 4 \), \( s=\frac{8}{3}\times16=\frac{128}{3}\approx42.67 \), point \( (4,\frac{128}{3}) \).
• When \( t = 5 \), \( s=\frac{8}{3}\times25=\frac{200}{3}\approx66.67 \), point \( (5,\frac{200}{3}) \).

Step3: Graph the points

Plot these points on a coordinate axis with the x - axis labeled "Time (t) in seconds" and the y - axis labeled "Distance (s) in feet", and draw a smooth curve (since it's a quadratic function) through them.

Answer:

24 feet

Part b