Question

Theo and Sasha start at the same place on a straight road riding bikes with the following velocities (measured in \(\mathrm{mi} / \mathrm{hr}\) ): Theo: \(v_{T}(t)=10,\) for \(t \geq 0\), Sasha: \(v_{S}(t)=15 t,\) for \(0 \leq t \leq 1\) and \(v_{S}(t)=15,\) for \(t>1\). a. Graph the velocity functions for both riders. b. If the riders ride for \(1 \mathrm{hr},\) who rides farther? Interpret your answer geometrically using the graphs of part (a). c. If the riders ride for 2 hr, who rides farther? Interpret your answer geometrically using the graphs of part (a). d. Which rider arrives first at the \(10-, 15\) -, and 20 -mile markers of the race? Interpret your answer geometrically using the graphs of part (a). e. Suppose Sasha gives Theo a head start of \(0.2 \mathrm{mi}\) and the riders ride for 20 mi. Who wins the race? f. Suppose Sasha gives Theo a head start of \(0.2 \mathrm{hr}\) and the riders ride for 20 mi. Who wins the race?

Short answer

Based on the given velocity functions, the following conclusions can be drawn: a. Theo's velocity function is a horizontal line at \(v_{T}(t)=10\), while Sasha's velocity function is piecewise with \(v_{S}(t)=15t\) for \(0 \leq t \leq 1\) and \(v_{S}(t)=15\) for \(t > 1\). b. Theo rides farther in 1 hour, covering 10 miles, while Sasha covers 7.5 miles. c. Sasha rides farther in 2 hours, covering 22.5 miles, while Theo covers 20 miles. d. Theo arrives first at the 10- and 15-mile markers, while Sasha arrives first at the 20-mile marker. e. With a 0.2-mile head start for Theo, Sasha wins the 20-mile race. f. With a 0.2-hour head start for Theo, Sasha still wins the 20-mile race.

Step-by-step solution

a. Graph the velocity functions for both riders.

To graph the velocity function for Theo, we have a simple horizontal line at \(v_{T}(t) = 10\) for all \(t \geq 0\). For Sasha, the velocity function has two parts: \(v_{S}(t) = 15t\) for \(0 \leq t \leq 1\) and \(v_{S}(t) = 15\) for \(t > 1\). This is a piecewise function that starts at the origin for \(t=0\), has a straight slope from \(0\) to \(15\) in the interval \([0, 1]\), and then becomes a horizontal line parallel to Theo's at \(v = 15\) for all \(t > 1\).

b. Who rides farther in 1 hr?

The distance each rider rides can be calculated by finding the area under their respective velocity graphs within the specified time interval, \(t=1\). The area under Theo's graph is a rectangle with base \(1\) and height \(10\), giving a total distance of \(10\) miles. For Sasha, the area under her graph in the first hour is a triangle with base \(1\) and height \(15\). The area of the triangle is \(0.5 \times 1 \times 15 = 7.5\) miles. Since Theo rides \(10\) miles and Sasha rides \(7.5\) miles in the first hour, Theo rides farther.

c. Who rides farther in 2 hr?

As in part b, we find the area under the velocity graphs for both riders within the time interval \(t=2\). For Theo, this remains a simple rectangle, now with a base of \(2\) (time) and a height of \(10\) (velocity). The total distance is \(2 \times 10 = 20\) miles. Sasha's distance in the 2-hour period consists of the same triangle (7.5 miles) as in part b, and the additional rectangle between \(t=1\) and \(t=2\) with base \(1\) and height \(15\). The area of this rectangle is \(1 \times 15 = 15\) miles. Adding the two distances for Sasha, we get \(7.5 + 15 = 22.5\) miles. In the first 2 hours, Theo rides \(20\) miles, and Sasha rides \(22.5\) miles, so Sasha rides farther.

d. Which rider arrives first at the 10-, 15-, and 20-mile markers?

To see who arrives first at each milestone, we look at when each rider covers that distance. With the velocity graphs, we look for the point in time when the area under the graph matches the milestone distance. - 10-mile marker: Theo covers 10 miles in 1 hour because his velocity is constant at \(10 \ \mathrm{mi}\ /\mathrm{hr}\). Sasha covers 10 miles when the area under her graph equals 10, which is between the first triangle and second rectangle of her velocity graph. Since the area of the first triangle (7.5 miles) is less than 10, we need to find the time when the rectangle area plus the triangle area equal 10: \(7.5 + 15(t-1) = 10\). Solving for \(t\), we get \(t = 1 + \frac{2.5}{15} = \frac{5}{3}\) or about \(1.67\) hours. Since Theo reaches 10 miles in 1 hour and Sasha takes about 1.67 hours, Theo arrives at the 10-mile marker first. - 15-mile marker: Theo covers 15 miles in \(t=\frac{3}{2}\) hours (1.5 hours) because \(10t = 15\). Sasha already covered 15 miles at \(t=1+1=2\) hours due to the triangle and the rectangle. Since Theo takes 1.5 hours to cover 15 miles and Sasha takes 2 hours to do so, Theo arrives at the 15-mile marker first. - 20-mile marker: Theo reaches 20 miles in 2 hours since his velocity remains constant at \(10 \ \mathrm{mi}\ /\mathrm{hr}\). By \(t=2\), Sasha has already covered 22.5 miles, as calculated in part c. Thus, Sasha reaches the 20-mile marker before 2 hours. Sasha arrives at the 20-mile marker before Theo.

e. Who wins the 20-mile race with a 0.2-mile head start for Theo?

We must first calculate the new time for Theo to cover 19.8 miles because of the given head start. Theo takes \(\frac{19.8}{10} = 1.98\) hours to cover the race. To find out when Sasha covers 20 miles, we can use the area under her graph once more. We already know she covers 22.5 miles in 2 hours, so she will need less time to cover 20 miles. Since 20 miles are between the 15-mile mark and the 22.5-mile mark, we need to find the time when the area covered from the start until that point equals 20. The area of the initial triangle plus the area of the rectangle between \(1 \leq t \leq x\) must equal 20: \(7.5 + 15(x-1) = 20\). Solving for \(x\), we get \(x = \frac{5}{6} + 1 = \frac{11}{6} \approx 1.83\) hours. Since Theo takes 1.98 hours to complete the race, and Sasha takes approximately 1.83 hours, Sasha wins the race.

f. Who wins the 20-mile race with a 0.2-hour head start for Theo?

With a 0.2-hour head start, Theo will cover \(10 \times 0.2 = 2\) miles during the head start. He will then need to cover the remaining 18 miles at his constant velocity of \(10\ \mathrm{mi}\ /\mathrm{hr}\). To do so, Theo will take \(\frac{18}{10} = 1.8\) hours to cover the remaining distance. In total, Theo takes \(0.2 + 1.8 = 2\) hours to complete the 20-mile race. As calculated previously, Sasha covers 20 miles in approximately 1.83 hours. Since Theo takes 2 hours to complete the race and Sasha takes approximately 1.83 hours, Sasha wins the race.

Key concepts

Graph Interpretation

When dealing with velocity functions, graphs can give us a visual representation of how speed changes over time. Analyzing these graphs helps in understanding different situations like acceleration or constant speed. In the case of Theo and Sasha, the velocity graph tells a story of their biking speeds.

• Theo's graph is a simple horizontal line at 10 mi/hr, indicating a steady pace throughout the journey.
• Sasha's graph is a piecewise function which initially shows acceleration and then stabilizes. This reflects her increasing speed initially, followed by a constant speed after a certain time.

Graph interpretation allows us to compare these two different biking patterns visually. By studying the area under each of their graphs, we can get valuable insights into who is biking faster over different intervals.

Piecewise Functions

Piecewise functions are crucial when a situation changes over time, like how Sasha's velocity changes. Here, Sasha’s function is divided into two distinct parts based on time:

• The first part is defined as \( v_{S}(t) = 15t \) for \( 0 \leq t \leq 1 \). This represents an initial acceleration where Sasha's speed increases linearly with time until \( t=1 \) hour.
• The second part is defined as \( v_{S}(t) = 15 \) for \( t > 1 \), showing that Sasha maintains a steady speed of 15 mi/hr post-acceleration.

Understanding such piecewise functions helps us interpret how a rider's speed strategy changes throughout the race, and how these changes affect distance traveled over time.

Distance Calculation

Calculating the distance using velocity functions involves finding the area under the velocity-time graph. This area represents the total distance covered. For constant velocities like Theo's, this is straightforward since we are dealing with rectangles. For example, to find how far Sasha rides in the first hour, we calculate the area under her velocity graph, which is a triangle during the first hour (\(7.5\) miles). After this initial hour, any computations would involve calculating rectangles.By adding areas of different shapes, like rectangles and triangles, we derive distances such as Sasha’s total of 22.5 miles in 2 hours, compared to Theo’s 20 miles over the same duration.

Head Start Analysis

A head start can change the dynamics of a race significantly. Whether in time or distance, it shifts when each rider needs to complete their 20-mile mark.

• Distance Head Start: If Theo gets a 0.2-mile advantage, the task involves calculating how soon Sasha closes this gap. Despite starting slightly behind, Sasha’s higher speed allows her to finish in about 1.83 hours while Theo takes 1.98 hours.
• Time Head Start: If Theo starts 0.2 hours earlier, he covers 2 miles before Sasha starts. Here too, because of her rapid pace, Sasha still finishes the race faster at 1.83 hours compared to Theo's 2 hours.

These analyses are useful to predict outcomes with different initial conditions, showing how and when Sasha manages to overtake Theo despite his head start.