Solutions Chapter 2 8
49. Slightly different answers may be obtained since the data comes from reading the graph.
(a) The greatest velocity is found at the highest point on the graph, which is at .
(b) The indication of a constant velocity on a velocity-time graph is a slope of 0, which occurs from
.
(c) The indication of a constant acceleration on a velocity-time graph is a constant slope, which
occurs from , again from , and again from .
(d) The magnitude of the acceleration is greatest when the magnitude of the slope is greatest, which
occurs from .
50. Slightly different answers may be obtained since the data comes from reading the graph.
(a) The instantaneous velocity is given by the slope of the tangent line to the curve. At ,
the slope is approximately .
(b) At , the slope of the tangent line to the curve, and thus the instantaneous velocity, is
approximately .
(c) The average velocity is given by .
(d) The average velocity is given by .
(e) The average velocity is given by .
51. Slightly different answers may be obtained since the data comes from reading the graph.
(a) The indication of a constant velocity on a position-time graph is a constant slope, which occurs
from .
(b) The greatest velocity will occur when the slope is the highest positive value, which occurs at
about .
(c) The indication of a 0 velocity on a position-time graph is a slope of 0, which occurs at about
from .
(d) When the slope is positive, from to ,
the object is moving in the positive direction. When the slope is negative, from to , the object is moving in the negative direction.
54. Slightly different answers may be obtained since the data comes from reading the graph.
(a) To estimate the distance the object traveled during the first minute, we need to find the area
under the graph, from t = 0 s to t = 60 s. Each "block" of the graph represents an "area" of . By counting and estimating, there are about 17.5 blocks under the 1st minute of the graph, and so the distance traveled during the 1st minute is about .
(b) For the second minute, there are about 5 blocks under the graph, and so the distance traveled
during the second minute is about .
Alternatively, average accelerations can be estimated for various portions of the graph, and then the
uniform acceleration equations may be applied. For instance, for part (a), break the motion up into
two segments, from 0 to 50 seconds and then from 50 to 60 seconds.
(a) t = 0 to 50:
55. The v vs. t graph is found by taking the slope of the x vs. t graph.
Both graphs are shown here.
56. (a) During the interval from A to B, it is , because its
displacement is negative.
(b) During the interval from A to B, it is , because the magnitude of its slope is
increasing (changing from less steep to more steep).
(c) During the interval from A to B, , because the graph is
concave downward, indicating that the slope is getting more negative, and thus the acceleration is negative.
(d) During the interval from D to E, it is , because the
displacement is positive.
(e) During the interval from D to E, it is , because the magnitude of its slope is
increasing (changing from less steep to more steep).
(f) During the interval from D to E, , because the graph is
concave upward, indicating the slope is getting more positive, and thus the acceleration
is positive.
(g) During the interval from C to D, .