18 worked answers · short steps first, explanations when you need them
Chapter 2 Exercises
Textbook pages 38–41
Question 1
For which motions can an object be treated as a point: a train between stations, a monkey riding a cyclist around a track, a sharply turning spinning cricket ball, or a beaker tumbling off a table?
Steps only
An object is a point object when its size and rotation do not matter for the motion being studied.
Answer
The train between stations and the monkey-cyclist on a large track can be treated as point objects. The cricket ball’s spin and sharp turn matter, and the tumbling beaker’s rotation matters, so those two cannot.
Question 2
Use the position-time graph of children A and B returning from school to compare (a) who lives nearer, (b) who starts earlier, (c) who walks faster, (d) whether they arrive together and (e) whether one overtakes the other.
Steps only
Read final positions for home distance, starting points for start time, line slopes for speed, and intersections for overtaking.
Answer
(a) A lives nearer. (b) A starts earlier. (c) B walks faster because B’s line is steeper. (d) They arrive at the same time. (e) B overtakes A once.
Question 3
A woman walks 2.5 km to work at 5 km/h from 9:00 am, stays until 5:00 pm, then returns by auto at 25 km/h. Give suitable scales and the key points for her position-time graph.
Steps only
Walking time = distance/speed = 2.5/5 = 0.5 h, so she reaches work at 9:30. Return time = 2.5/25 = 0.1 h = 6 min.
Answer
Plot time horizontally and distance from home vertically. Join (9:00, 0) to (9:30, 2.5 km), draw a horizontal segment to (5:00 pm, 2.5 km), then join to (5:06 pm, 0). Any clear convenient scale is acceptable.
Question 4
A person takes 5 one-metre steps forward and 3 one-metre steps backward, one step per second. Starting 13 m from a pit, how long until he reaches it? Describe the position-time graph.
Steps only
One full pattern takes 8 s and gives a net gain of 2 m. After four patterns (32 s), he is 8 m from the start.
Answer
He needs 5 more forward steps to reach 13 m, taking 5 s. Total = 32+5 = 37 s. The graph is a repeating zigzag: up 5 m in 5 s, down 3 m in 3 s, until it reaches 13 m.
Question 5
A car travels at 126 km/h and stops in 200 m with uniform retardation. Find its retardation and stopping time.
Steps only
Convert 126 km/h to 35 m/s. Use v²=u²+2as with v=0, u=35 m/s and s=200 m.
Answer
0=35²+2a(200), so a=−3.06 m/s². Retardation magnitude = 3.06 m/s². Then v=u+at gives t=35/3.06≈11.4 s.
Question 6
A ball is thrown up at 29.4 m/s. State the acceleration direction and the velocity/acceleration at the top. With the top as x=0 and downward positive, give signs during the trip. Find the maximum height and total return time (g=9.8 m/s²).
Steps only
Gravity points downward throughout. At the top, velocity is momentarily zero but acceleration is still 9.8 m/s² downward. With downward positive and origin at the top, the ball is below the top both on the way up and down, so x is positive; v is negative going up and positive coming down; a stays positive.
Answer
Height = u²/(2g)=29.4²/(19.6)=44.1 m. Time up = u/g=3 s, so total return time = 6 s.
Question 7
For one-dimensional motion, decide whether each statement is true: (a) zero speed can occur with non-zero acceleration; (b) zero speed can occur with non-zero velocity; (c) constant speed requires zero acceleration; (d) positive acceleration always means speeding up.
Steps only
Check the definitions: speed is |velocity|, while acceleration describes how velocity changes. The sign of acceleration depends on the chosen positive direction.
Answer
(a) True: at the top of a vertical throw, speed is zero but gravity still accelerates the ball. (b) False: zero speed means zero velocity. (c) True for one-dimensional motion at constant speed: velocity cannot change direction without first reaching zero. (d) False: positive acceleration can slow an object moving in the negative direction.
Question 8
A ball is dropped from 90 m. At each floor collision it loses one-tenth of its speed. Describe the speed-time graph from 0 to 12 s (use g=9.8 m/s²).
Steps only
During each free flight, speed changes at 9.8 m/s each second. The first fall time is √(2h/g)=√(180/9.8)≈4.29 s and impact speed is gt≈42.0 m/s.
Answer
At impact the speed drops to 0.9×42=37.8 m/s. It rises linearly to zero after 37.8/9.8≈3.86 s, then falls linearly back to 37.8 m/s at t≈12.0 s. Draw straight sloping segments with a sudden downward jump at the first impact.
Question 9
Explain the difference between displacement and path length, and between average velocity magnitude and average speed. Show which is greater or equal, and when they are equal.
Steps only
Displacement depends only on start and finish; path length counts the whole route. The shortest route between two positions cannot exceed the path actually travelled.
Answer
Thus path length ≥ |displacement|. Dividing both by the same time gives average speed ≥ |average velocity|. Equality holds when motion stays along one straight line in one direction without turning back.
Question 10
A person walks 2.5 km to a market at 5 km/h, then returns at 7.5 km/h. Find |average velocity| and average speed for (i) first 30 min, (ii) full 50 min, and (iii) first 40 min.
Steps only
(i) At 30 min he is 2.5 km from home: displacement and path are both 2.5 km in 0.5 h. (ii) At 50 min he is home: displacement 0, path 5 km, time 5/6 h. (iii) In the extra 10 min after reaching market he returns 1.25 km, so he is 1.25 km from home; path is 3.75 km in 2/3 h.
Answer
(i) |average velocity|=2.5/0.5=5 km/h; average speed=5 km/h.
(ii) |average velocity|=0; average speed=5/(5/6)=6 km/h.
(iii) |average velocity|=1.25/(2/3)=1.875 km/h; average speed=3.75/(2/3)=5.625 km/h.
Question 11
Why is instantaneous speed always equal to the magnitude of instantaneous velocity, although average speed and average velocity magnitude may differ?
Steps only
Over a finite time, a moving object may change direction, making the path longer than its net displacement. Over an infinitesimally short time, the small path segment has the same length as the magnitude of the small displacement.
Four proposed graphs show (a) one position at two times for a single instant, (b) two velocities at one time, (c) negative speed, and (d) decreasing total path length. Which cannot represent one-dimensional motion?
Steps only
Position and velocity must each have a single value at each time. Speed cannot be negative, and total distance travelled cannot decrease.
Answer
All four are impossible: (a) and (b) assign multiple values at one time; (c) makes speed negative; (d) makes total path length go down.
Question 13
An x-t graph is straight for t<0 and curved like a parabola for t>0. Does that mean the object changes from a straight path to a parabolic path? Suggest a physical example.
Steps only
An x-t graph plots position along one axis against time; its curve is not a drawing of the route through space.
Answer
No. One example is an object at rest until t=0, then released to fall vertically under gravity. It still moves along one straight vertical line.
Question 14
A police van moving at 30 km/h fires a bullet forward at 150 m/s relative to the van. A thief’s car ahead moves in the same direction at 192 km/h. At what speed does the bullet hit the car?
Steps only
Convert vehicle speeds: van = 30×5/18=8.33 m/s; car = 192×5/18=53.33 m/s. Bullet speed relative to ground = 150+8.33=158.33 m/s.
Answer
Closing speed = 158.33−53.33 = 105 m/s. The bullet hits at 105 m/s relative to the car.
Question 15
For each graph in the textbook figure, suggest a physical event that could produce its changing motion.
Steps only
Match each graph’s variable and shape to a physical change: position shows location, velocity shows signed rate of motion, and acceleration shows how velocity changes.
Answer
A suitable example for a position graph that repeatedly rises and falls with smaller changes is a bouncing ball losing height. A velocity graph with sudden reversals can represent a ball rebounding from a surface. A brief acceleration spike can represent the short collision when a ball is hit.
Question 16
From the simple-harmonic-motion x-t graph, state signs of x, v and a at t=0.3 s, 1.2 s and −1.2 s.
Steps only
Position is read from whether the curve is above or below x=0; velocity is the slope; in SHM, acceleration has the opposite sign to position (a=−ω²x).
Answer
At 0.3 s: x<0, slope<0 so v<0, and a>0. At 1.2 s: x>0, slope>0 so v>0, and a<0. At −1.2 s: x<0, slope>0 so v>0, and a>0.
Question 17
The x-t graph has three equal time intervals. Identify where average speed is largest and smallest, and give the average-velocity sign in each interval.
Steps only
On an x-t graph, slope gives velocity. The magnitude of slope gives speed; its sign gives direction.
Answer
Average speed is greatest in interval 3 and least in interval 2. Average velocity is positive in interval 1, negative in interval 2, and positive in interval 3.
Question 18
A speed-time graph has three equal intervals and marked points A–D. Identify the interval with greatest |average acceleration| and greatest average speed; give signs of velocity and acceleration in each interval and accelerations at A–D.
Steps only
The steepest graph segment has the greatest acceleration magnitude. The highest average graph value gives the greatest average speed. Since motion is in the chosen positive direction, velocity is positive throughout; acceleration follows whether speed rises or falls.
Answer
|average acceleration| is greatest in interval 2; average speed is greatest in interval 3. Signs: interval 1 v>0, a>0; interval 2 v>0, a<0; interval 3 v>0, a≈0. At A, B, C and D the tangent is horizontal, so instantaneous acceleration is 0.