Q#51
(II) Figure 2–42 shows the velocity of a train as a function of time.
(a) At what time was its velocity greatest?
(b) During what periods, if any, was the velocity constant?
(c) During what periods, if any, was the acceleration constant?
(d) When was the magnitude of the acceleration greatest?
Answer:
Slightly different answers may be obtained since the data come from reading the graph.
(a) The greatest velocity is found at the highest point on the graph, which is at t ≈ 48 s
(b) The indication of a constant velocity on a velocity vs. time graph is a slope of 0, which occurs from t = 90 s to t ≈ 108 s
(c) The indication of a constant acceleration on a velocity vs. time graph is a constant slope, which occurs from t = 0 s to t ≈ 42 s , again from t ≈ 65 s to t ≈ 83 s , and again from t = 90 s to ≈ 108 s.
(d) The magnitude of the acceleration is greatest when the magnitude of the slope is greatest, which occurs from t ≈ 65 s to t ≈ 83 s.
Q#52
(II) A sports car accelerates approximately as shown in the velocity–time graph of Fig. 2–43. (The short flat spots in the curve represent manual shifting of the gears.) Estimate the car’s average acceleration in
(a) second gear and
(b) fourth gear.
Answer:
Slightly different answers may be obtained since the data come from reading the graph. We assume that the short, nearly horizontal portions of the graph are the times that shifting is occurring, and those times are not counted as being “in” a certain gear.
(a) The average acceleration in 2nd gear is given by
$\overline{a}_2=\frac{\Delta v_2}{\Delta t_2}$
$\overline{a}_2=\frac{24 \ m/s - 14 \ m/s}{8 \ s - 4 \ s}=2.5 \ m/s^2$
(b) The average acceleration in 4th gear is given by
$\overline{a}_4=\frac{\Delta v_4}{\Delta t_4}$
$\overline{a}_4=\frac{44 \ m/s - 37 \ m/s}{27 \ s - 16 \ s}=0.64 \ m/s^2$
Q#53
(II) The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2–44. What is its instantaneous velocity
(a) at t = 10.0 s and
(b) at t = 30.0 s.
What is its average velocity
(c) between t = 0 and t = 5.0 s
(d) between t = 25.0 s and t = 30.0 s and
(e) between t = 40.0 s and t = 50.0 s?
Answer:
Slightly different answers may be obtained since the data come from reading the graph.
(a) The instantaneous velocity is given by the slope of the tangent line to the curve. At t = 10.0 s, the slope is approximately
$v(10) \approx \frac{3 \ m -0}{10.0 \ s - 0}=0.30 \ m/s$
(b) At t = 30.0 s, the slope of the tangent line to the curve, and thus the instantaneous velocity, is approximately
$v(30) \approx \frac{20 \ m -8 \ m}{35 \ s - 25 \ s}=1.2 \ m/s$
(c) The average velocity is given by
$\overline{v} = \frac{x(5) -x(0)}{5.0 \ s - 0 \ s}=\frac{1.50 \ m -0}{5.0 \ s}=0.30 \ m/s$
(d) The average velocity is given by
$\overline{v} = \frac{x(30) -x(25)}{30.0 \ s - 25.0\ s}=\frac{16.0 \ m -9.0 \ m}{5.0 \ s}=1.4 \ m/s$
(e) The average velocity is given by
$\overline{v} = \frac{x(50) -x(40)}{50.0 \ s - 40.0 \ s}=\frac{10.0 \ m -19.5 \ m}{10.0 \ s}=-0.95 \ m/s$
Q#54
(II) In Fig. 2–44,
(a) during what time periods, if any, is the velocity constant?
(b) At what time is the velocity greatest?
(c) At what time, if any, is the velocity zero?
(d) Does the object move in one direction or in both directions during the time shown?
Answer:
Slightly different answers may be obtained since the data come from reading the graph.
(a) The indication of a constant velocity on a position versus time graph is a constant slope, which occurs from t = 0 s to t ≈ 18 s.
(b) The greatest velocity will occur when the slope is the highest positive value, which occurs at about t = 27 s
(c) The indication of a 0 velocity on a position versus time graph is a slope of 0, which occurs at about t = 38 s.
(d) The object moves in both directions. When the slope is positive, from t = 0 s to t = 38 s, the object is moving in the positive direction.
When the slope is negative, from t = 38 s to t = 50 s, the object is moving in the negative direction.
Q#55
(III) Sketch the v vs. graph for the object whose displacement as a function of time is given by Fig. 2–44.
Answer:
The v vs. t graph is found by taking the slope of the x vs. t graph. Both graphs are shown here.
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