Problem#1
A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t) = bt2 – ct3 where b = 2.40 m/s2 and c = 0.120 m/s3. (a) Calculate the average velocity of the car for the time interval t = 0 to t = 10.0 s. (b) Calculate the instantaneous velocity of the car at t = 0, t = 5.0 s and t = 10.0 s. (c) How long after starting from rest is the car again at rest?Answer:
x(t) = bt2 – ct3 where b = 2.40 m/s2 and c = 0.120 m/s3, then
x(t) = (2.40 m/s2)t2 – (0.120 m/s3)t3
at t = 0; x = 0.
(a) with x(10.0 s) = (2.40 m/s2)(10.0 s)2 – (0.120 m/s3)(10.0 s)3 = 120 m
the average velocity of the car for the time interval t = 0 to t = 10.0 s is
vavg = ∆x/∆t = 120 m/10.0 s = 12.0 m/s
(b) the instantaneous velocity of the car at t = 0, t = 5.0 s and t = 10.0 s
vx = dx/dt = (4.80 m/s)t – (0.360 m/s)t2
at t = 0: vx = 0
at t = 5.0 s; vx = (4.80 m/s)(5.0 s) – (0.360 m/s)(5.0 s)2 = 15.0 m/s
at t = 10.0 s; vx = (4.80 m/s)(10.0 s) – (0.360 m/s)(10.0 s)2 = 12.0 m/s
(c) long after starting from the rest is the car again at the rest, when vx = 0,
vx = dx/dt = (4.80 m/s2)t – (0.360 m/s3)t2
0 = t(4.80 – 0.0360t)
t = 0 and t = 13.3 s
Problem#2
A bird is flying due east. Its distance from a tall building is given by x(t) = 28.0 m + (24.0 m/s)t – (0.0450 m/s3)t3. What is the instantaneous velocity of the bird when t = 8.00 s ?
Answer:
We know the position x(t) of the bird as a function of time and want to find its instantaneous velocity at a particular time.
The instantaneous velocity is
vx = dx/dt = d[28.0 m + (24.0 m/s)t – (0.0450 m/s3)t3]/dt
vx = (12.4 m/s) – (0.135 m/s3)t2
then,
at t = 8.0 s, vx = (12.4 m/s) – (0.135 m/s3)(8.0 s)2 = 3.76 m/s
Problem#3
A ball moves in a straight line (the x-axis). The graph in Fig. 1 shows this ball’s velocity as a function of time. (a) What are the ball’s average speed and average velocity during the first 3.0 s? (b) Suppose that the ball moved in such a way that the graph segment after 2.0 s was –3 m/s instead of +3 m/s. Find the ball’s average speed and average velocity in this case.
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| Fig.1 |
(a) For t = 0 to t = 2.0 s, Δx = (2.0 m/s)(2.0 s) = 4.0 m.
For t = 2.0 s to t = 3.0 s, Δx = (3.0 m/s)(1.0 s) = 3.0 m.
For the first 3.0 s, ∆x = 4.0 m + 3.0 m = 7.0 m
The distance traveled is also 7.0 m
The average velocity is
vavg = ∆x/∆t = 7.0 m/3.0 s = 2.33 m/s
The average speed is also 2.33 m/s.
(b) For t = 2.0 s to t = 3.0 s, Δx = (–3.0 m/s)(1.0 s) = –3.0 m.
For the first 3.0 s, Δx = 4.0 m + (–3.0 m) = +1 m
The dog runs 4.0 m in the +x-direction and then 3.0 m in the −x-direction, so the distance traveled is still 7.0 m
The average velocity is
vavg = ∆x/∆t = 1.0 m/3.0 s = 0.33 m/s
The average speed is
Savg = 7.00 m/3.00 s = 2.33 m/s
Problem#4
A physics professor leaves her house and walks along the sidewalk toward campus. After 5 min it starts to rain and she returns home. Her distance from her house as a function of time is shown in Fig. 2. At which of the labeled points is her velocity (a) zero? (b) constant and positive? (c) constant and negative? (d) increasing in magnitude? (e) decreasing in magnitude?
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| Fig.2 |
The instantaneous velocity is the slope of the tangent to the x versus t graph.
(a) The velocity is zero where the graph is horizontal; point IV
(b) The velocity is constant and positive where the graph is a straight line with positive slope; point I
(c) The velocity is constant and negative where the graph is a straight line with negative slope; point V
(d) The slope is positive and increasing at point II. (e) The slope is positive and decreasing at point III
The sign of the velocity indicates its direction.
Problem#5
A test car travels in a straight line along the x-axis. The graph in Fig. 3 shows the car’s position x as a function of time. Find its instantaneous velocity at points A through G.
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| Fig.3 |
Find the instantaneous velocity of a car using a graph of its position as a function of time.
The instantaneous velocity at any point is the slope of the x versus t graph at that point. Estimate the slope from the graph.
at points A; vx = dx/dt = (40 m – 20 m)/(3 s – 0) = +6.7 m/s
at points B; vx = dx/dt = (40 m – 20 m)/(3 s – 0) = +6.7 m/s
at points C; vx = dx/dt = (40 m – 40 m)/(5 s – 3 s) = 0 m/s
at points D; vx = dx/dt = (0 m – 40 m)/(5 s – 4 s) = –40.0 m/s
at points E; vx = dx/dt = (0 m – 40 m)/(5 s – 4 s) = –40.0 m/s
at points F; vx = dx/dt = (0 m – 40 m)/(5 s – 4 s) = –40.0 m/s
at points G; vx = dx/dt = 0
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