Problem#1
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.Answer:
(a) Taking derivatives of x(t) = 12t2 – 2t3 we obtain the velocity and the acceleration functions:
v(t) = 24t – 6t2 and a(t) = 24 – 12t
with length in meters and time in seconds. Plugging in the value t = 3 yields x(3) = 54 m
(b) Similarly, plugging in the value t = 3 yields v(3) = 18 m/s.
(c) For t = 3, a(3) = –12 m/s2
(d) At the maximum x, we must have v = 0; eliminating the t = 0 root, the velocity equation reveals t = 24/6 = 4 s for the time of maximum x. Plugging t = 4 into the equation for x leads to x = 64 m for the largest x value reached by the particle.
(e) From (d), we see that the x reaches its maximum at t = 4.0 s.
(f) A maximum v requires a = 0, which occurs when t = 24/12 = 2.0 s. This, inserted into the velocity equation, gives vmax = 24 m/s.
(g) From (f), we see that the maximum of v occurs at t = 24/12 = 2.0 s.
(h) In part (e), the particle was (momentarily) motionless at t = 4 s. The acceleration at that time is readily found to be 24 – 12(4) = –24 m/s2 .
(i) The average velocity is defined by Eq. 2-2, so we see that the values of x at t = 0 and t = 3 s are needed; these are, respectively, x = 0 and x = 54 m (found in part (a)). Thus,
vavg = ∆x/∆t = (54 m – 0)/(3 s – 0) = 18 m/s
Problem#2
At a certain time a particle had a speed of 18 m/s in the positive x direction, and 2.4 s later its speed was 30 m/s in the opposite direction. What is the average acceleration of the particle during this 2.4 s interval?
Answer:
We represent the initial direction of motion as the +x direction. The average acceleration over a time interval 1 2 t tt ≤ ≤ is given by
aavg = ∆v/∆t
Let v1 = +18 m/s at t1 = 0 and v2 = –30 m/s at t2 = 2.4 s.
we find
aavg = (–30 m/s – 18 m/s)/(2.4 s – 0) = –20 m/s2
The average acceleration has magnitude 20 m/s2 and is in the opposite direction to the particle’s initial velocity. This makes sense because the velocity of the particle is decreasing over the time interval.
Problem#3
(a) If the position of a particle is given by x = 20t – 5t3 , where x is in meters and t is in seconds, when, if ever, is the particle’s velocity zero? (b) When is its acceleration a zero? (c) For what time range (positive or negative) is a negative? (d) Positive? (e) Graph x(t), v(t), and a(t).
Answer:
We use the functional notation x(t), v(t) and a(t) and find the latter two quantities by differentiating:
v(t) = dx/dt = 20 – 15t2 and a(t) = dv/dt = –30t
with SI units understood. These expressions are used in the parts that follow.
the particle’s velocity zero is
0 = 20 – 15t2
t = 1.2 s
(a) its acceleration a zero (a(t) = 0),
(b) From 0 = – 30t, we find a(0) = 0 (that is, it vanishes at t = 0).
(c) It is clear that a(t) = – 30t is negative for t > 0
(d) The acceleration a(t) = – 30t is positive for t < 0
(e) The graphs are shown Fig.1 below. SI units are understood.
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| Fig.1 |
From t = 0 to t = 5.00 min, a man stands still, and from t = 5.00 min to t = 10.0 min, he walks briskly in a straight line at a constant speed of 2.20 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 2.00 min to 8.00 min? What are (c) vavg and (d) aavg in the time interval 3.00 min to 9.00 min? (e) Sketch x versus t and v versus t, and indicate how the answers to (a) through (d) can be obtained from the graphs.
Answer:
We use Eq. average velocity and Eq. average acceleration. Regarding our coordinate choices, the initial position of the man is taken as the origin and his direction of motion during 5 min ≤ t ≤ 10 min is taken to be the positive x direction. We also use the fact that Δx = vΔt' when the velocity is constant during a time interval Δt'.
(a) The entire interval considered is Δt = 8 – 2 = 6 min, which is equivalent to 360 s, whereas the sub-interval in which he is moving is only Δt' = 8 − 5 = 3 min = 180 s. His position at t = 2 min is x = 0 and his position at t = 8 min is x = vΔt′ = (2.2)(180) = 396 m. Therefore,
vavg = ∆x/∆t = (396 m – 0)/(360 s) = 1.10 m/s
(b) The man is at rest at t = 2 min and has velocity v = +2.2 m/s at t = 8 min. Thus, keeping the answer to 3 significant figures,
aavg = ∆v/∆t = (2.2 m/s – 0)/(360 s) = 0.00611 m/s2
(c) Now, the entire interval considered is Δt = 9 – 3 = 6 min (360 s again), whereas the sub-interval in which he is moving is Δt′ = 9 − 5 = 4 min = 240 s). His position at t = 3 min is x = 0 and his position at t = 9 min is x = v Δt′ = (2.2)(240) = 528 m . Therefore,
vavg = ∆x/∆t = (528 m – 0)/(360 s) = 1.47 m/s
(d) The man is at rest at t = 3 min and has velocity v = +2.2 m/s at t = 9 min. Consequently, aavg = 2.2/360 = 0.00611 m/s2 just as in part (b).
(e) The horizontal line near the bottom of this x-vs-t graph (Fig.2) represents the man standing at x = 0 for 0 ≤ t < 300 s and the linearly rising line for 300 ≤ t ≤ 600 s represents his constant-velocity motion. The lines represent the answers to part (a) and (c) in the sense that their slopes yield those results.
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| Fig.2 |
The graph of v-vs-t is not shown here, but would consist of two horizontal “steps” (one at v = 0 for 0 ≤ t < 300 s and the next at v = 2.2 m/s for 300 ≤ t ≤ 600 s). The indications of the average accelerations found in parts (b) and (d) would be dotted lines connecting the “steps” at the appropriate t values (the slopes of the dotted lines representing the values of aavg).
Problem#5
The position of a particle moving along the x axis depends on the time according to the equation x = ct2 – bt3 , where x is in meters and t in seconds.What are the units of (a) constant c and (b) constant b? Let their numerical values be 3.0 and 2.0, respectively. (c) At what time does the particle reach its maximum positive x position? From t = 0.0 s to t = 4.0 s, (d) what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s. Find its acceleration at times (j) 1.0 s, (k) 2.0 s, (l) 3.0 s, and (m) 4.0 s.
Answer:
In this solution, we make use of the notation x(t) for the value of x at a particular t. The notations v(t) and a(t) have similar meanings.
(a) Since the unit of ct2 is that of length, the unit of c must be that of length/time2 , or m/s2 in the SI system.
(b) Since bt3 has a unit of length, b must have a unit of length/time3, or m/s3 .
(c) When the particle reaches its maximum (or its minimum) coordinate its velocity is zero. Since the velocity is given by
v = dx/dt = 2ct – 3bt2
v = 0 occurs for
t(2c – 3bt) = 0
t = 0 or t = 2c/3b = 2(3.0 m/s2)/(3 x 2.0 m/s3) = 1.0 s
For t = 0, x = x0 = 0 and for t = 1.0 s, x = 1.0 m > x0. Since we seek the maximum, we reject the first root (t = 0) and accept the second (t = 1s).
(d) In the first 4 s the particle moves from the origin to x = 1.0 m, turns around, and goes back to
x = ct2 – bt3
x(4 s) = (3.0 m/s2)(4 s)2 – (2.0 m/s3)(4.0 s)3 = –80 m
The total path length it travels is 1.0 m + 1.0 m + 80 m = 82 m.
(e) Its displacement is Δx = x2 – x1, where x1 = 0 and x2 = –80 m. Thus, Δx = −80 m .
The velocity is given by
v = 2ct – 3bt2 = (6.0 m/s2)t – (6.0 m/s3)t2
(f) Plugging in t = 1 s, we obtain
v(1 s) = (6.0 m/s2)(1.0 s) – (6.0 m/s3)(1.0 s)2 = 0
(g) Similarly,
v(2 s) = (6.0 m/s2)(2.0 s) – (6.0 m/s3)(2.0 s)2 = –12 m/s
(h) v(3 s) = (6.0 m/s2)(3.0 s) – (6.0 m/s3)(3.0 s)2 = –36 m/s
(i) v(4 s) = (6.0 m/s2)(4.0 s) – (6.0 m/s3)(4.0 s)2 = –72 m/s
The acceleration is given by
a = dv/dt = 2c – 6bt = 6.0 m/s2 – (12.0 m/s3)t
(j) Plugging in t = 1 s, we obtain
a(1 s) = 6.0 m/s2 – (12.0 m/s3)(1.0 s) = –6.0 m/s2
(k) a(2 s) = 6.0 m/s2 – (12.0 m/s3)(2.0 s) = –18.0 m/s2
(l) a(3 s) = 6.0 m/s2 – (12.0 m/s3)(3.0 s) = –30.0 m/s2
(m) a(4 s) = 6.0 m/s2 – (12.0 m/s3)(4.0 s) = –42.0 m/s2
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