Question

example 2.2 slowly moving train goal obtain average and instantaneous velocities from a graph. problem a train moves slowly along a straight portion of track according to the graph of position versus time in the left - hand figure. (a) find the average velocity for the total trip. (b) find the average velocity during the first 4.00 s of motion. (c) find the average velocity during the next 4.00 s of motion. (d) find the instantaneous velocity at t = 2.00 s (e) find the instantaneous velocity at t = 9.00 s. strategy the average velocities can be obtained by substituting the data into the definition. the instantaneous velocity at t = 2.00 s is the same as the average velocity at that point because the position vs. time graph is a straight line, indicating constant velocity. finding the instantaneous velocity when t = 9.00 s requires sketching a line tangent to the curve at that point and finding its slope. solution (a) find the average velocity from ⓐ to ⓒ calculate the slope of the dashed blue

Explanation:

Step1: Recall average - velocity formula

The average - velocity formula is $v_{avg}=\frac{\Delta x}{\Delta t}$, where $\Delta x=x_f - x_i$ and $\Delta t=t_f - t_i$.

Step2: (a) For total trip

From the graph, assume the initial position $x_i = 0$ m at $t_i = 0$ s and the final position $x_f=10$ m at $t_f = 12$ s. Then $\Delta x=10 - 0=10$ m and $\Delta t = 12-0 = 12$ s. So, $v_{avg}=\frac{10 - 0}{12-0}=\frac{10}{12}=\frac{5}{6}\approx0.83$ m/s.

Step3: (b) First 4.00 s

At $t_i = 0$ s, $x_i = 0$ m and at $t_f = 4$ s, $x_f = 4$ m. Then $\Delta x=4 - 0 = 4$ m and $\Delta t=4 - 0=4$ s. So, $v_{avg}=\frac{4 - 0}{4 - 0}=1$ m/s.

Step4: (c) Next 4.00 s (from $t = 4$ s to $t = 8$ s)

At $t_i = 4$ s, $x_i = 4$ m and at $t_f = 8$ s, $x_f = 4$ m. Then $\Delta x=4 - 4 = 0$ m and $\Delta t=8 - 4 = 4$ s. So, $v_{avg}=\frac{4 - 4}{8 - 4}=0$ m/s.

Step5: (d) Instantaneous velocity at $t = 2.00$ s

Since the position - time graph is a straight line from $t = 0$ s to $t = 4$ s, the instantaneous velocity is the same as the average velocity in this interval. From step 3, $v = 1$ m/s.

Step6: (e) Instantaneous velocity at $t = 9.00$ s

Sketch a tangent line to the curve at $t = 9$ s. The tangent line passes through two points (assume $(8,4)$ and $(10,8)$). The slope of the tangent line (instantaneous velocity) is $v=\frac{8 - 4}{10 - 8}=2$ m/s.

Answer:

(a) $\frac{5}{6}\approx0.83$ m/s
(b) 1 m/s
(c) 0 m/s
(d) 1 m/s
(e) 2 m/s