Explanation:
a) The dependent variable is the velocity of the scooter, and the independent variable is time.
b) Using the kinematic equation v = u + at, where u is the initial velocity, a is the acceleration, t is time, and v is the final velocity, we can find the velocity as a function of time for t ≤ 6 seconds:
v = u + at = 1.0 + (a × t)
c) The graph of velocity as a function of time for the scooter would be a straight line with a positive slope, starting at (0,1.0) and ending at (6,4.0). The x-axis would be labeled "Time (s)" and the y-axis would be labeled "Velocity (mph)".
d) We can find the acceleration of the scooter by rearranging the equation from part (b) to a = (v - u) / t, where v = 4.0 mph, u = 1.0 mph, and t = 6 seconds:
a = (4.0 - 1.0) / 6 = 0.5 mph/s
We can label the acceleration as 0.5 mph/s on the graph.
e) The initial velocity of the scooter is 1.0 mph, which we can label on the graph.
f) To find the distance traveled by the scooter between 0 and 6 seconds using the graphical method, we can calculate the area under the velocity-time graph. The graph forms a trapezoid with a base of 6 seconds, a height of 3 mph (the difference between the final and initial velocities), and a top length of 1.0 mph. The distance traveled is therefore:
distance = 0.5 × (1.0 + 4.0) × 6 = 15.0 miles
g) To find the distance traveled by the scooter between 0 and 6 seconds using the kinematic equations, we can use the equation s = ut + 0.5at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is time. Since the acceleration is constant, we can use the average velocity (2.5 mph) to find the distance traveled:
s = 1.0 × 6 + 0.5 × 0.5 × 6^2 = 15.75 miles
The result from the kinematic equation is slightly larger than the result from the graphical method. This is likely due to rounding errors and the approximation of the trapezoidal area under the velocity-time graph. Overall, both methods provide a good estimate of the distance traveled by the scooter between 0 and 6 seconds.