Physics 140
HOMEWORK Chapter 2A
Q1. Figure 2-14 gives the velocity of a particle moving on an x-axis. What are (a) the initial and
(b) the final directions of travel? (c) Does the particle stop momentarily? (d) Is the acceleration
positive or negative? (e) Is it constant or varying?
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(a) Negative. The initial value of v is below the zero line.
(b) Positive. The final value of v is above the zero line.
(c) Yes. When the line crosses the t-axis, v = 0.
(d) Positive. The slope of v vs t is positive.
(e) Constant. The plot of v is a line, meaning constant slope.
Q3. Figure 2-16 shows four paths along which objects move from a starting point to a final point, all
in the same time interval. The paths pass over a grid of equally spaced straight lines. Rank the paths
according to (a) the average velocity of the objects and (b) the average speed of the objects, greatest
first
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(a) All paths have the same average velocity: each shows a NET displacement of one grid line to the
right, and they were all accomplished in equal time intervals.
(b) No 4 > No 1 = No 2 > No 3. Since the time intervals are equal, speeds rank in order of total path
length.
Q4. Figure 2-17 is a graph of a particle’s position along an x-axis versus time. (a) At time t = 0, what
is the sign of the particle’s position? Is the particle’s velocity positive, negative, or 0 at (b) t = 1 s,
(c) t = 2 s, and (d) t = 3 s? (e) How many times does the particle go through the point x = 0?
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(a) Position is Negative; x(0) is below the t axis.
(b) Velocity is Positive at t = 1 s because slope is upward.
(c) Velocity is zero at t = 2 s.
(d) Velocity is Negative at t = 3 s.
(e) Twice: at t = 1 s and t = 3 s.
Q7. Hanging over the railing of a bridge, you drop an egg (no initial velocity) as you throw a second
egg downward. Which curves in Fig. 2-19 give the velocity v(t) for (a) the dropped egg and (b) the
thrown egg? (Curves A and B are parallel; so are C, D, and E; so are F and G.)
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(a) Curve D. (All the curves are straight lines.) The initial velocity must be zero, and velocity must
decrease (have an increasing magnitude in the negative direction).
(b) Curve E. Initial velocity must be negative, and velocity must become more negative at time passes.
Curve G is eliminated by the fact that the acceleration of the thrown egg is equal to that of the dropped
egg. Curve G has a greater (in magnitude) acceleration.
P3. An automobile travels on a straight road for 40 km at 30 km/h. It then continues in the same
direction for another 40 km at 60 km/h. (a) What is the average velocity of the car during the full 80km trip? (Assume that it moves in the positive x direction.) (b) What is the average speed? (c) Graph
x versus t and indicate how the average velocity is found on the graph.
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(a) The ONLY way to do this type of problem is to work it out one step at a time. Let t1 and t2 be
the respective times of the two legs. Also, leave everything in km and hours.
t1 = (40 km)/(30 km/hr) = 1.333 hr t2 = (40 km)/(60 km/hr) = 0.6667 hr ttotal = 2.0 hr
vavg = (80 km)/(2 hr) = +40 km/hr.
(b) Average speed is (80 km)/2 hr) = 40 km/hr. In this case, average speed is apparently equal to
average velocity. The difference would be that velocity is specifically positive (direction). They are
equal numerically only because all the motion was in the same direction.
(c) Line from (0 hr, 0 km) to (1.33 hr, 40 km) to 2 hr, 80 km). Average velocity is the slope of the line
connecting the first and last points.
P6. The 1992 world speed record for a bicycle (human-powered vehicle) was set by Chris Huber. His
time through the measured 200 m stretch was a sizzling 6.509 s, at which he commented, “Cogito ergo
zoom!” (I think, therefore I go fast!). In 2001, Sam Whittingham beat Huber’s record by 19.0 km/h.
What was Whittingham’s time through the 200 m?
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We need to determine Huber’s speed in km/hr, then add 19.0 for Whittingham’s, then calculate Whittingham’s time. Use subscripts H and W. Also, you should determine easily that 1 m/s = 3.6 km/hr.
vH = (200 m)/(6.509 s) = 30.73 m/s = 110.6 km/hr.
vW = 129.6 km/hr = 36.00 m/s.
tW = (200 m)/(36.00 m/s) = 5.556 s.
P13. You drive on Interstate 10 from San Antonio to Houston, half the time at 55 km/h and the
other half at 90 km/h. On the way back you travel half the distance at 55 km/h and the other half
at 90 km/h. What is your average speed (a) from San Antonio to Houston, (b) from Houston back
to San Antonio, and (c) for the entire trip? (d) What is your average velocity for the en- tire trip?
(e) Sketch x versus t for (a), assuming the motion is all in the positive x direction. Indicate how the
average velocity can be found on the sketch.
——— Recognize a slightly complicated distance-rate-time problem. Leave units in km and hr. Use 4
sigdig (55.00 and 90.00) to emphasize the technique.
(a) Notice that we aren’t given the distance, nor the time. Do we need them? The only way is to
work it out. Let d = distance between the two cities; and t0 be the total time SA to H. Further, let
v1 = 55 km/hr, v2 = 90 km/hr, and d1 and d2 be the distances traveled at each speed. Note that
d = d1 + d2 . Then:
d1 = (v1 )(t0 /2).
d2 = (v2 )(t0 /2).
Add the equations to get d1 + d2 = d = (v1 + v2 )t0 /2, or
|v| = d/t0 = (v1 + v2 )/2 = 72.50 km/hr.
(b) This time let t1 and t2 be the times, each leg having a distance of d/2. t1 + t2 = t3 , where t3 is the
total time for the return. The value of t3 will not be the same as t0 .
d/2 = v1 t1 , or t1 = d/(2v1 ).
d/2 = v2 t2 , or t2 = d/(2v2 ). Add the times to get
t1 + t2 = t3 = (d/2)(1/v1 + 1/v2 ). Get the average as
|vavg | = d/t3 = 2/(1/v1 + 1/v2 ) = 2v1 v2 /(v1 + v2 ).
Plug numbers to get
|vavg | = 68.28 km/hr.
(c) The legs SA to H and H to SA are the same distance. We can apply the result of part (b),
vavg = 2v1 v2 /(v1 + v2 ), but with v1 = 72.50 km/hr and v2 = 68.28 km/hr.
|v|avg = 2v1 v2 /(v1 + v2 ) = 70.32 km/hr.
(d) The average velocity is ZERO, since the DISPLACEMENT for the round trip is zero..
(e) Similar to P.3(c) above.
P18. The position of a particle moving along an x-axis is given by x = 12t2 − 2t3 , where x is in meters
and t is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle
at t = 3.0 s. (d) What is the maximum positive coordinate reached by the particle and (e) at what
time is it reached? (f) What is the maximum positive velocity reached by the particle and (g) at what
time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving
(other than at t = 0)? (i) Determine the average velocity of the particle between t = 0 and t = 3 s.
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We have
x(t) = 12t2 − 2t3 .
Use v(t) = dx/dt to get
v(t) = 24t − 6t2 .
Use a(t) = dv/dt to get
a(t) = 24 − 12t.
Note that the constants include units which are not written out, but implied by the requirement that
t be in sec and x in meters.
(a) x(3 s) = 12 · (32 ) − 2 · (33 ) = 108 − 54 = 54 m.
(b) v(3 s) = 24 · (31 ) − 6 · (32 ) = 72 − 54 = 18 m/s.
(c) a(3 s) = 24 − 12 · (31 ) = 24 − 36 = −12 m/s2 .
(d) By inspection of the equation for v(t), it is a parabola concave-down. We recall that the extreme
position occurs when v(t) = 0, or
24t − 6t2 = 0 ⇒ t = {0, 4 s}. Evaluate both possibilities.
x(0) = 0 and x(4 s) = 64 m. So, choose the second value:
Maximum value of x is 64 m. We are assuming that the motion starts at t = 0, and that negative
values for t are irrelevant.
(e) The max displacement occurs at t = 4 s, per part (d)
(f) The extreme velocity occurs when dv/dt = 0, or a(t) = 0.
24 − 12t = 0 ⇒ t = 2 s
v(2 s) = 24 · 2 − 6 · 22 = 24 m/s.
(g) Max positive velocity at t = 2 s, per part (f).
(h) “The particle is not moving” ⇒ v = 0, or
24t − 6t2 = 0 ⇒ t = 4 s.
a(4 s) = 24 − 12 · (4) = −24 m/s2 .
(i) Note first that acceleration is NOT constant, so we can’t use the constant-acceleration equations.
We need the original definition:
vavg = ∆x/∆t = [x(3 s) − x(0)]/(3 s) = [54 m − 0]/(3 s) = +18 m/s.
P25. An electric vehicle starts from rest and accelerates at a rate of 2.0 m/s2 in a straight line until it
reaches a speed of 20 m/s. The vehicle then slows at a constant rate of 1.0 m/s2 until it stops. (a) How
much time elapses from start to stop? (b) How far does the vehicle travel from start to stop?
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(a) Let t1 be the time at a = a1 = +2 m/s2 and t2 be the time at a = a2 = −1 m/s2 .
During the a1 phase v(t) = vi + a1 t ⇒ t1 = (vf − vi )/a1 = (20 m/s − 0)/(2 m/s2 ) = 10 s.
Since we took a1 as positive, we take a2 as negative: a2 = −1 m/s2 . During the a2 phase v(t) =
vi + a2 t ⇒ t2 = (vf − vi )/a2 = (0 − 20 m/s)/(−1 m/s2 ) = 20 s.
Total time = t1 + t2 = (10 s + 20 s) = 30 s
(b) Use d1 and d2 for the DISPLACEMENTS during the respective phases.
d1 = vi t1 + (1/2)a1 t21 = 0 + (1/2)(2 m/s2 )(10 s)2 = 100 m.
d2 = vi t2 + (1/2)a2 t22 = (+20 m/s)(20 s) + (1/2)(−1 m/s2 )(20 s)2 = 400 m − 200 m = 200 m.
Total distance = (100 m + 200 m = 300 m.
P31. Suppose a rocket ship in deep space moves with constant acceleration equal to 9.8 m/s2 , which
gives the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to
acquire a speed one-tenth that of light, which travels at 3.0 × 108 m/s? (b) How far will it travel in so
doing?
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Recognize a “constant acceleration” problem and apply the constant acceleration equations, noting
that vf = 0.1 × 3 × 108 m/s = 3 × 107 m/s.
(a) v(t) = vi + at ⇒ t = (v − vi )/a = (3 × 107 m/s − 0)/(9.8 m/s2 ) = 3.06 × 106 s (35 days)
(b) ∆x = vi t + (1/2)at2 = 0 + (1/2)(9.8 m/s2 )(3.06 × 106 s)2 = 4.59 × 1013 m.
(I am keeping one more sigdig than justified to help with math checking.) For comparison, the distance
to Proxima Centauri (the nearest star other than the Sun) is about 4 × 1016 m, while the distance to
Neptune is about 4.5 × 1012 m. So, this rocket ship is about 10 times as far as the outermost planet
(Pluto has been declared a non-planet), but 1/1000 the way to the next star (if headed in the right
direction).