A group of skydivers gains speed as they fall together.

Question

Six people hold hands to form a ring while falling.
A graph relating the group's kinetic energy in \[(\text{kJ})\] and their speed in meters per second

\[\left(\tfrac{\text{m}}{\text{s}}\right)\] is shown below.
A graph with kinetic energy in kilojoules on the y-axis and speed in meters per second on the x-axis. Starting from the bottom of the y-axis, the increments are labeled 0, 20, 40, 60, 80. Starting from the left end of the x-axis, the increments are labeled 0, 5, 10, 15, 20.
A curved line starts at point (0,0) and curves up and to the right, ending at point (20,80). The shape of the curve looks like the right side of a bowl.
Use the pattern on the graph to complete the prediction.
When the group’s speed is

\[40\,\tfrac{\text{m}}{\text{s}},\] their kinetic energy will be
\[\text{kJ}.\]

Answer 1

To predict the kinetic energy of the group when their speed is \(40,\tfrac{\text{m}}{\text{s}}\), we can analyze the pattern in the graph.

Given the curve starts at the origin (0,0) and reaches (20,80), we observe the trend indicates that the kinetic energy increases as the speed increases. The relationship between kinetic energy \(KE\) and speed \(v\) is typically quadratic, expressed mathematically as:

\[ KE = \frac{1}{2} mv^2 \]

In the context of this question, we can observe the graph's relationship. Let's check the values we already know from the graph:

• At \(v = 0,\tfrac{\text{m}}{\text{s}}\), \(KE = 0,\text{kJ}\)
• At \(v = 10,\tfrac{\text{m}}{\text{s}}\), the kinetic energy is \(40,\text{kJ}\)
• At \(v = 20,\tfrac{\text{m}}{\text{s}}\), the kinetic energy is \(80,\text{kJ}\)

Now we can find the kinetic energy at \(v = 40,\tfrac{\text{m}}{\text{s}}\). Since the graph suggests a quadratic increase, it seems likely that when the speed doubles (from \(20,\tfrac{\text{m}}{\text{s}}\) to \(40,\tfrac{\text{m}}{\text{s}}\)), the kinetic energy also increases more dramatically based on the pattern.

Given that at \(20,\tfrac{\text{m}}{\text{s}}\) (where \(KE = 80,\text{kJ}\)) we can expect another doubling of kinetic energy at \(40,\tfrac{\text{m}}{\text{s}}\). Hence we can assume:

\[ KE (40,\tfrac{\text{m}}{\text{s}}) = 4 \times KE (10,\tfrac{\text{m}}{\text{s}}) = 4 \times 40,\text{kJ} = 160,\text{kJ} \]

Thus, the predicted kinetic energy when the group’s speed is \(40,\tfrac{\text{m}}{\text{s}}\) will be:

\[ \boxed{160} \]

So, the answer is \(160,\text{kJ}\).