Problem#1
The position of a pinewood derby car was observed at various times; the results are summarized in the following table. Find the average velocity of the car for (a) the first second, (b) the last 3 s, and (c) the entire period of observation.Answer:
(a) the average velocity of the car for the first second is
vavg = ∆x/∆t = 2.3 m/1.0 s = 2.3 m/s
(b) the average velocity of the car for the last 3 s is
vavg = ∆x/∆t = (57.5 m – 9.2 m)/3.0 s = 16.1 m/s
(c) the average velocity of the car for the entire period of observation is
vavg = ∆x/∆t = (57.5 m – 0)/5.0 s = 11.5 m/s
Problem#2
(a) Sand dunes in a desert move over time as sand is swept up the windward side to settle in the lee side. Such “walking” dunes have been known to walk 20 feet in a year and can travel as much as 100 feet per year in particularly windy times. Calculate the average speed in each case in m/s. (b) Fingernails grow at the rate of drifting continents, on the order of 10 mm/yr. Approximately how long did it take for North America to separate from Europe, a distance of about 3 000 mi?
Answer:
(a) the average speed in each case in m/s is
vavg = ∆x/∆t = (20 ft/1 yr)(1 m/3.281 ft)(1 yr/3.156 x 107 s) = 2 x 10-7 m/s or in particularly windy times
vavg = ∆x/∆t = (100 ft/1 yr)(1 m/3.281 ft)(1 yr/3.156 x 107 s) = 1 x 10-6 m/s
(b) The time required must have been
∆t = ∆x/vavg
∆t = (3000 mi/10 mm/yr)(1609 m/1 mi)(103 mm/1m) = 5 x 108 yr
Problem#3
The position versus time for a certain particle moving along the x axis is shown in Figure 1. Find the average velocity in the time intervals (a) 0 to 2 s, (b) 0 to 4 s, (c) 2 s to 4 s, (d) 4 s to 7 s, (e) 0 to 8 s.
 |
| Fig.1 |
Answer:
(a) the average velocity in the time intervals 0 to 2 s is
vavg = ∆x/∆t = 10 m/2 s = 5 m/s
(b) the average velocity in the time intervals 0 to 4 s is
vavg = ∆x/∆t = 5 m/4 s = 1.2 m/s
(c) the average velocity in the time intervals 2 to 4 s is
vavg = ∆x/∆t = (5 m – 10 m)/(4 s – 2 s) = –2.5 m/s
(d) the average velocity in the time intervals 4 to 7 s is
vavg = ∆x/∆t = (–5 m – 5 m)/(7 s – 4 s) = –3.3 m/s
(d) the average velocity in the time intervals 0 to 8 s is
vavg = ∆x/∆t = (0 – 0)/(8 s – 0) = 0 m/s
Problem#4
A particle moves according to the equation x = 10t2 where x is in meters and t is in seconds. (a) Find the average velocity for the time interval from 2.00 s to 3.00 s. (b) Find the average velocity for the time interval from 2.00 to 2.10 s.
Answer:
the equation x = 10t2;
(a) the average velocity for the time interval from 2.00 s to 3.00 s is
For t = 2.00 s, x = 10(2.00 s)2 = 40.0 m
For t = 3.00 s, x = 10(3.00 s)2 = 90.0 m
Then,
vavg = ∆x/∆t = 50 m/1.0 s = 50.0 m/s
(a) the average velocity for the time interval from 2.00 s to 2.10 s is
For t = 2.10 s, x = 10(2.10 s)2 = 44.1 m
Then,
vavg = ∆x/∆t = 4.1 m/0.1 s = 41.0 m/s
Problem#5
A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00 m/s. What is (a) her average speed over the entire trip? (b) her average velocity over the entire trip?
Answer:
(a) Let d represent the distance between A and B. Let t1 be the time for which the walker has the higher speed in
5.00 m/s = d/t1
Let t2 represent the longer time for the return trip in
–3.00 m/s = –d/t2
The average speed is
Savg = total distance/total time
Savg = (d + d)/[(d/5.00 m/s) + (d/3.00 m/s)] = 3.75 m/s
(b) She starts and finishes at the same point A. With total displacement = 0, average velocity = 0
Problem#6
A block on a nearly frictionless table is attached to a spring (Fig. 1). The other end of the spring is fixed to a wall. When the block is pulled to the right and then released, it moves back and forth for a long time before coming to rest.
 |
| Fig.2 |
 |
| Table.1 |
(a) Create a position-versus-time graph from these data, (b) what is the displacement of the block during the entire 6.25 s?, (c) what is the displacement of the block during the first 3.25 s?, (d) what is the total distance traveled for the recorded measurement interval?
Answer:
(a) Make a position-versus-time graph with z on the vertical axis and t on the horizontal axis (Fig. 3)
 |
| Fig.3 |
(b) In 6.25 s, the block returned to the original position, so we expect the displacement to be zero.
The initial and final positions are at the same height on the graph. The displacement of the block during the entire 6.25 s is
∆z = zf – zi = (0.65k m – 0.65k m) = 0
(c) the displacement of the block during the first 3.25 s is
∆z = zf – zi = (–0.65k m – 0.65k m) = –1.30k m
(d) the total distance traveled for the recorded measurement interval is
From part C, we know that the block moved 1.30 m in the first 3.25 s. In the next 3 s, it returned to its original position, so it must have traveled another 1.30 m. The total distance traveled is thus the sum of these two distances that is
d = (1.30 m + 1.30 m) = 2.60 m
Problem#7
After a long and grueling race, two cadets, A and B, are coming into the finish line at the Marine Corps marathon. They move in the same direction along a straight path; the positionversus- time graphs for the runners are shown in Figure 4 for a minute near the end of the race. (a). At time t1 is the speed of cadet B greater than, less than, or equal to the speed of cadet A? (b). At time t2, is cadet B speeding up, slowing down, or moving with constant speed? (c). From time t = 0 to time t = 60 s, is the average speed of cadet B greater than, less than, or equal to the average speed of cadet A? Explain.
 |
| Fig.4 |
Answer:
(a) At time t1 is the speed of cadet B less than the speed of cadet A.
(b) At time t2, is cadet B speeding up.
(c). From time t = 0 to time t = 60 s, is the average speed of cadet B less than the average speed of cadet A, less than because the slope of the line drawn from t = 0 to 60 s for B is less than the slope of the similar line for A
Post a Comment for "Position, Velocity, and Speed Problems and Solutions"