Question 2.2: Slowly Moving Train GOAL Obtain average and instantaneous ve...
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 Figure 2.7a. Find (a) the average velocity for the total trip, (b) the average velocity during the first 4.00 s of motion, (c) the average velocity during the next 4.00 s of motion, (d) the instantaneous velocity at t = 2.00 s, and (e) 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.
(a) Find the average velocity from ⓞ to © :
Calculate the slope of the dashed blue line:
\bar{v} = \frac{Δx}{Δt} = \frac{10.0 m}{12.0 s} = +0.833 m/s(b) Find the average velocity during the first 4 seconds of the train’s motion.
Again, find the slope:
\bar{v} = \frac{Δx}{Δt} = \frac{4.00 m}{4.00 s} = +1.00 m/s(c) Find the average velocity during the next 4 seconds.
Here, there is no change in position as the train moves from Ⓐ to Ⓑ , so the displacement Δx is zero:
\bar{v} = \frac{Δx}{Δt} = \frac{0 m}{4.00 s} = 0 m/s(d) Find the instantaneous velocity at t = 2.00 s.
This is the same as the average velocity found in (b), because the graph is a straight line: v = 1.00 m/s
(e) Find the instantaneous velocity at t = 9.00 s.
The tangent line appears to intercept the x-axis at (3.0 s, 0 m) and graze the curve at (9.0 s, 4.5 m). The instantaneous velocity at t = 9.00 s equals the slope of the tangent line through these points:
\bar{v} = \frac{Δx}{Δt} = \frac{4.5 m – 0 m}{9.0 s – 3.0 s} = 0.75 m/sREMARKS From the origin to Ⓐ, the train moves at constant speed in the positive x-direction for the first 4.00 s, because the position vs. time curve is rising steadily toward positive values. From Ⓐ to Ⓑ, the train stops at x = 4.00 m for 4.00 s. From Ⓑ to ©, the train travels at increasing speed in the positive x-direction.