Problem #1
(a) Is it possible to move with constant speed but not constant velocity?(b) Is it possible to move with constant velocity but not constant speed?
Answer:
a.) Yes. For example, if a car travels at constant speed while going around a curve in the road, its speed remains constant. But since velocity also includes direction, and the car's direction is changing, the velocity is not constant.
b.) No. Since speed is the magnitude of velocity, any object with constant velocity must have constant speed.
Problem #2
A girl walks 10 km in 4 hours and 4 km in 1 hour in the same direction.
(a) What is the girl's average speed for the whole journey?
(b) What is the girl's average velocity for the whole journey?
Answer:
Given: s1 = 10 km, t1 = 4 hours and s2 = 4 km, t2 = 1 hours, and travel Fig.1 of girls are shown below
 |
| Fig.1 |
(a) average speed = distance/time = (10 km + 4 km)/(4 h + 1 h) = 2.8 km/h
(b) average velocity = displacement/time = (10 km + 4 km)/(4 h + 1 h) = 2.8 km/h
Problem #3
A girl walks 10 km East in 4 hours and then 4 km West in 1 hour.
(a) What is the girl's average speed for the whole journey?
(b) What is the girl's average velocity for the whole journey?
Answer:
Given: s1 = 10 km, t1 = 4 hours and s2 = 4 km, t2 = 1 hours, and travel Fig.2 of girls are shown below
(b) average velocity = displacement/time = (10 km + 4 km)/(4 h + 1 h) = 2.8 km/h
Problem #3
A girl walks 10 km East in 4 hours and then 4 km West in 1 hour.
(a) What is the girl's average speed for the whole journey?
(b) What is the girl's average velocity for the whole journey?
Answer:
Given: s1 = 10 km, t1 = 4 hours and s2 = 4 km, t2 = 1 hours, and travel Fig.2 of girls are shown below
 |
| Fig.2 |
(a) average speed = distance/time = (10 km + 4 km)/(4 h + 1 h) = 2.8 km/h
(b) average velocity = displacement/time = (10 km – 4 km)/(4 h + 1 h) = 1.2 km/h
Problem #4
A car takes 20 s to move around a round about of radius 14 m. Calculate:
(a) average speed, and
(b) magnitudo pf average velocity
Answer:
Given: time t = 20 s and radius r = 14 m.
(a) average speed is the ratio of total distance to total time taken. So, distance convered in one revolution
s = 2πr = 2 x 22/7 x 14 m = 88 m
average speed = s/t = 88/20 = 4,4 m/s
(b) in one complete revolution displacement of car is zero.
vavr = displacement/time = 0/10 = 0 m/s
Problem #5
a train travels from city A to city B with a constant speed of 20 ms-1 and return back to city A with a constant speed of 30 m.s-1. Find its average speed during te entire journey.
Answer:
Given: vA = 20 m.s-1 and vB = 30 m.s-1.
Consider, the distance between the two cities A and B = x m.
Time taken by the train to travel from A to B = x/20 = t1 (say)
Time taken to come back from B to A = x/30 (say)
So,
average speed = total distance/total time = (x + x )/(t1 + t2)
= 2x/[(x/20) + (x/30)] = 2 x 20/3
average speed = 40/3 m.s-1
Problem #6
Richard swam and took a 50 m swimming pool for 20 s. Then, he turned around and returned to his starting position in 22 s. Calculate Richard's average speed at:
(a) the first part of the trip (towards the opposite)
(b) the second part of the trip (back)
(c) the whole journey (going and returning)
Answer:
Given: s1 = 50 m, t1 = 20 s, and s2 = 50 m, t2 = 22 s,
(a) the first part of the trip (towards the opposite)
Average speed = displacement/time = 50/20 = 2.5 m.s-1
(b) the second part of the trip (back)
Average speed = displacement/time = –50/20 = 2.5 m.s-1
(c) the whole journey (going and returning)
Average speed = displacement/time = (50 – 50)/42 = 0 m.s-1
Problem #7
A runner ran 6 km to the north, then 8 km to the east. Note the runner's time is 2 hours. Determine:
(a) distance and displacement
(b) average speed and average velocity.
Answer:
Given: s1 = 6 km to north, s2 = 8 km to east, total time t = 2 h. The runner's route is shown in Fig.3,
(b) average velocity = displacement/time = (10 km – 4 km)/(4 h + 1 h) = 1.2 km/h
Problem #4
A car takes 20 s to move around a round about of radius 14 m. Calculate:
(a) average speed, and
(b) magnitudo pf average velocity
Answer:
Given: time t = 20 s and radius r = 14 m.
(a) average speed is the ratio of total distance to total time taken. So, distance convered in one revolution
s = 2πr = 2 x 22/7 x 14 m = 88 m
average speed = s/t = 88/20 = 4,4 m/s
(b) in one complete revolution displacement of car is zero.
vavr = displacement/time = 0/10 = 0 m/s
Problem #5
a train travels from city A to city B with a constant speed of 20 ms-1 and return back to city A with a constant speed of 30 m.s-1. Find its average speed during te entire journey.
Answer:
Given: vA = 20 m.s-1 and vB = 30 m.s-1.
Consider, the distance between the two cities A and B = x m.
Time taken by the train to travel from A to B = x/20 = t1 (say)
Time taken to come back from B to A = x/30 (say)
So,
average speed = total distance/total time = (x + x )/(t1 + t2)
= 2x/[(x/20) + (x/30)] = 2 x 20/3
average speed = 40/3 m.s-1
Problem #6
Richard swam and took a 50 m swimming pool for 20 s. Then, he turned around and returned to his starting position in 22 s. Calculate Richard's average speed at:
(a) the first part of the trip (towards the opposite)
(b) the second part of the trip (back)
(c) the whole journey (going and returning)
Answer:
Given: s1 = 50 m, t1 = 20 s, and s2 = 50 m, t2 = 22 s,
(a) the first part of the trip (towards the opposite)
Average speed = displacement/time = 50/20 = 2.5 m.s-1
(b) the second part of the trip (back)
Average speed = displacement/time = –50/20 = 2.5 m.s-1
(c) the whole journey (going and returning)
Average speed = displacement/time = (50 – 50)/42 = 0 m.s-1
Problem #7
A runner ran 6 km to the north, then 8 km to the east. Note the runner's time is 2 hours. Determine:
(a) distance and displacement
(b) average speed and average velocity.
Answer:
Given: s1 = 6 km to north, s2 = 8 km to east, total time t = 2 h. The runner's route is shown in Fig.3,
 |
| Fig.3 |
(a) distance = s1 + s2 = 6 + 8 = 14 km
Displacement = [(6)2 + (8)2]1/2 = 10 km
(b) average speed = distance/time = 14 km/2h = 7.0 km.h-1
Average velocity = displacement/time = 10 km/2h = 5.0 km.h-1
Displacement = [(6)2 + (8)2]1/2 = 10 km
(b) average speed = distance/time = 14 km/2h = 7.0 km.h-1
Average velocity = displacement/time = 10 km/2h = 5.0 km.h-1
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