Question

exercise
hints: getting started | im stuck!
use the values from practice it to help you work this exercise. figure (b) above graphs another run of the train.
(a) find the average velocity from ⓐ to ⓒ.
if an object travels a net distance of 0, at what average velocity did it travel? m/s
(b) find the average and instantaneous velocity from ⓐ to ⓓ.
average
your response differs from the correct answer by more than 10%. double check your calculations. m/s
instantaneous your response differs from the correct answer by more than 10%. double check your calculations. m/s
(c) find the approximate instantaneous velocity at t = 7.0 s.
your response differs from the correct answer by more than 10%. double check your calculations. m/s
(d) find the average velocity on the open interval from ⓑ to ⓒ and instantaneous velocity at t = 9.9 s.
average
the response you submitted has the wrong sign. m/s
instantaneous
the response you submitted has the wrong sign. m/s

Explanation:

Step1: Recall average - velocity formula

The average - velocity formula is $v_{avg}=\frac{\Delta x}{\Delta t}$, where $\Delta x$ is the net displacement and $\Delta t$ is the time interval. If the net distance (assuming displacement here as we are dealing with velocity) is 0, then $v_{avg}=\frac{0}{\Delta t}=0$.

Step2: For average and instantaneous velocity from 0 to A

To find the average velocity from 0 to A, we use $v_{avg}=\frac{x_A - x_0}{t_A - t_0}$. To find the instantaneous velocity at a point, we consider the slope of the tangent line to the position - time graph at that point. If the position - time graph is a straight line from 0 to A, the average velocity and instantaneous velocity are equal and given by the slope of the line connecting the two points.

Step3: For instantaneous velocity at t = 7.0 s

The instantaneous velocity at a particular time $t$ is the slope of the tangent line to the position - time graph at $t$. We estimate this slope by looking at the local behavior of the graph near $t = 7.0$s.

Step4: For average and instantaneous velocity from B to C

The average velocity from B to C is $v_{avg}=\frac{x_C - x_B}{t_C - t_B}$. The instantaneous velocity at $t = 9.9$s is the slope of the tangent line to the position - time graph at $t = 9.9$s.

Answer:

(a) 0
(b - average) and (b - instantaneous): Need the position - time values of 0 and A to calculate.
(c) Need the position - time graph to estimate the slope of the tangent at t = 7.0 s.
(d - average): Need the position - time values of B and C to calculate. (d - instantaneous): Need the position - time graph to estimate the slope of the tangent at t = 9.9 s.