Question

Two hours after the start of a 100-kilometer bicycle race, a cyclist passes the 2 kilometer mark while riding at a velocity of 43 kilometers per hour. Complete parts ( through (C) below (A) Find the cyclists average velocity during the first two hours of the race kilometers per hour (8) 100 represent the distance traveled (n kilometers) from the start of the race (x0) to time x in hours) Find the slope of the secant ine through the points 100) and (2.1211 on the graph of y 100 kometers per hour (C) Find the equation of the tangent line to the graph of y-fox) at the point (2.12) Type an equation

Answer 1

The cyclist's average velocity during the first two hours of the race is 1 kilometer per hour. The slope of the secant line through the points (0,0) and (2,121) is 60.5 kilometers per hour. The equation of the tangent line to the graph of y = f(x) at the point (2,12) is y = 60.5x - 109.

Step-by-step explanation:

(A) To find the cyclist's average velocity during the first two hours of the race, we need to find the total distance traveled and divide it by the time taken. The cyclist passes the 2 kilometer mark after 2 hours, so the total distance traveled is 2 kilometers. Therefore, the average velocity is 2 kilometers divided by 2 hours, which is 1 kilometer per hour.

(B) To find the slope of the secant line through the points (0,0) and (2,121), we need to find the change in the y-coordinate divided by the change in the x-coordinate. The change in y-coordinate is 121 - 0 = 121 kilometers per hour, and the change in x-coordinate is 2 - 0 = 2 hours. Therefore, the slope of the secant line is 121 kilometers per hour divided by 2 hours, which is 60.5 kilometers per hour.

(C) The equation of the tangent line to the graph of y = f(x) at the point (2,12) can be found using the point-slope form of a linear equation. The slope of the tangent line is the same as the slope of the secant line, which is 60.5 kilometers per hour. So, using the point-slope form with the point (2,12), we have: y - 12 = 60.5(x - 2). Simplifying this equation gives: y = 60.5x - 109.

Answer 2

(A) The cyclist's average velocity during the first two hours of the race is 50 kilometers per hour.

(B) The slope of the secant line through the points (0,0) and (2,100) is 50 kilometers per hour.

(C) The equation of the tangent line to the graph of (y = 50x) at the point (2,100) is (y = 50x).

(A) To find the average velocity, we use the formula \( \text{Average Velocity} =
`\frac{\text{Change in Distance}}{\text{Change in Time}} \).` In this case, the cyclist covers 100 kilometers in the first two hours, resulting in an average velocity of
`\( \frac{100 \, \text{km}}{2 \, \text{hours}} = 50 \, \text{km/h} \).`

(B) The slope of the secant line between two points on the graph of the function represents the average rate of change. In this scenario, the points (0,0) and (2,100) are used to calculate the slope. The formula for slope (m) is
`\( \frac{\text{Change in } y}{\text{Change in } x} \), resulting in \( (100)/(2) = 50 \, \text{km/h} \).`

(C) The equation of the tangent line at a given point on a graph can be determined using the slope-intercept form (y = mx + b), where (m) is the slope and (b) is the y-intercept. Since the slope is 50 (as found in part B), and the point of tangency is (2,100), the equation of the tangent line is (y = 50x). This reflects the instantaneous velocity of the cyclist at the specific time (2 hours) on the graph.