Question 2.2: SLOWLY MOVING TRAIN GOAL Obtain average and instantaneous ve...

(a) Find the average velocity from Ⓞ to ©.

Calculate the slope of the dashed blue line:

\bar{v}=\frac{\Delta x}{\Delta t}=\frac{10.0 \mathrm{~m}}{12.0 \mathrm{~s}}=+0.833 \mathrm{~m} / \mathrm{s}

(b) Find the average velocity during the first 4 seconds of the train’s motion.

Again, find the slope:

\bar{v}=\frac{\Delta x}{\Delta t}=\frac{4.00 \mathrm{~m}}{4.00 \mathrm{~s}}=+1.00 \mathrm{~m} / \mathrm{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 \Delta x is zero:

\bar{v}=\frac{\Delta x}{\Delta t}=\frac{0 \mathrm{~m}}{4.00 \mathrm{~s}}=0 \mathrm{~m} / \mathrm{s}

(d) Find the instantaneous velocity at t=2.00 \mathrm{~s}.

This is the same as the average velocity found in (b), because the graph is a straight line:

v=1.00 \mathrm{~m} / \mathrm{s}

(e) Find the instantaneous velocity at t=9.00 \mathrm{~s}.

The tangent line appears to intercept the x-axis at (3.0 \mathrm{~s}, 0 \mathrm{~m}) and graze the curve at (9.0 \mathrm{~s}, 4.5 \mathrm{~m}). The instantaneous velocity at t=9.00 \mathrm{~s} equals the slope of the tangent line through these points:

v=\frac{\Delta x}{\Delta t}=\frac{4.5 \mathrm{~m}-0 \mathrm{~m}}{9.0 \mathrm{~s} \quad 3.0 \mathrm{~s}}=0.75 \mathrm{~m} / \mathrm{s}

REMARKS From the origin to Ⓐ , the train moves at constant speed in the positive x direction for the first 4.00 \mathrm{~s}, because the position vs. time curve is rising steadily toward positive values. From Ⓐ to Ⓑ, the train stops at x=4.00 \mathrm{~m} for 4.00 \mathrm{~s}. From Ⓑ to © , the train travels at increasing speed in the positive x direction.