Velocity and Speed: Solutions to Problems



            There is no denying the fact that the rate at which the body changes its position in a definite direction is called velocity. Velocity has both magnitude and direction for which it is called vector quantity. On the other hand, speed has only magnitude for which it is known as scalar quantity. In this context I will try to show the distinctions between speed and velocity by setting forth the problems with which the crucial and analytical problem may be solved.
Problem 1: A man walks 7 km in 2 hours and 2 km in 1 hour in the same direction.

a) What is the man's average speed for the whole journey?

b) What is the man's average velocity for the whole journey?
Solution to Problem 1:

a)
speed and velocity - Problem 1
average speed =  
distance
time
=
7 km + 2 km
2 hours + 1 hour
=
9 km
3 hours
= 3 km/h
b)
average velocity =  
displacement
time
=
7 km + 2 km
2 hours + 1 hour
=
9 km
3 hours
= 3 km/h
Problem 2: A man walks 7 km East in 2 hours and then 2.5 km West in 1 hour.

a) What is the man's average speed for the whole journey?

b) What is the man's average velocity for the whole journey?
Solution to Problem 2:

a)
speed and velocity - Problem 2
average speed =  
distance
time
=
7 km + 2.5 km
2 hours + 1 hour
=
9.5 km
3 hours
= 3.2 km/h (approximated to the nearest tenth)
b)
average velocity =  
displacement
time
=
7 km - 2.5 km
2 hours + 1 hour
=
4.5 km
3 hours
= 1.5 km/h
average speed =  
distance
time
=  
circumference
time
=
2 * 0.5 * Pi
1 hour
= Pi km/h = 3.14 km/h (approximated)
b) If you walk around a circular field and come back to the same point where you started the displacement, which a change in position, is equal to zero. Since the displacement is equal to zero, the average velocity is also equal to zero.
Problem 4: John drove South 120 km at 60 km/h and then East 150 km at 50 km/h. Determine

a) the average speed for the whole journey?

b) the magnitude of the average velocity for the whole journey?
Solution to Problem 4:

a)
speed and velocity - Problem 4
The time t1 to cover 120 km at a speed of 60 km/h is given by

t1 = 120 / 60 = 2 hours

The time t2 to cover 150 km at a speed of 50 km/h is given by

t2 = 150 / 50 = 3 hours
average speed =  
distance
time
=
120 km + 150 km
2 hours + 3 hours
=
270 km
5 hours
= 54 km/h
b) The magnitude of the displacement is the distance AC between the final point and the starting point and is calculated using Pythagora's theorem as follows

AC2 = 1202 + 1502 AC = √(14400+22500) = 30 √41 km
average velocity =  
displacement
time
=
30 √41 km
2 hours + 3 hour
= 38.4 km/h (approximated)
Problem 5: If I can walk at an average speed of 5 km/h, how many miles I can walk in two hours?
Solution to Problem 5:

distance = (average speed) * (time) = 5 km/h * 2 hours = 10 km

using the rate of conversion 0.62 miles per km, the distance in miles is given by

distance = 10 km * 0.62 miles/km = 6.2 miles
Problem 6: A train travels along a straight line at a constant speed of 60 mi/h for a distance d and then another distance equal to 2d in the same direction at a constant speed of 80 mi/h.

a)What is the average speed of the train for the whole journey?
Solution to Problem 6:

a) The time t1 to cover distance d at a speed of 60 mi/h is given by

t1 = d / 60

The time t2 to cover distance 2d at a speed of 80 mi/h is given by

t2 = 2d / 80
average speed =  
distance
time
=
d + 2d
d/60 + 2d/80
=
3d
(80d + 2d*60)/(60*80)
=
3d
200d/4800
=
3d*4800
200d
= 72 mi/h
Problem 7: A car travels 22 km south, 12 km west, and 14 km north in half an hour.

a) What is the average speed of the car?

b) What is the final displacement of the car?

c) What is the average velocity of the car?
Solution to Problem 7:

a)
speed and velocity - Problem 1
average speed =  
distance
time
=
22 km + 12 km + 14 km
0.5 hour
= 96 km/h
average velocity =  
displacement
time
=
4√13 km
0.5 hour
= 28.8 km/h (approximated)
velocity and speed - Problem 8
a) the average speed, in m/s, for the whole journey.

b) the magnitude of the displacement.

c) the magnitude of the average velocity, in m/s, for the whole journey.
Solution to Problem 8:

a)
velocity and speed - Problem 8
average speed =  
distance
time
=
3 km + 1 km + 1.5 km + 0.5 km + 0.5 km
3250 seconds
=
6.5 km
3250 seconds
=
6500 m
3250 seconds
= 2 m/s
b) The magnitude of the displacement is the distance between the starting point A and the final point F and is the hypotenuse AF of the right triangle AFH and is calculated using Pythagoras’s theorem as follows

AF2 = FH2 + HA2 = 32 + 42 = 9 + 16 = 25

DA = √25 = 5 km

c)
average velocity =  
displacement
time
=
5 km
3250 s
=
5000 m
3250 s
= 1.5 m/s (approximated)

In determining the speed and velocity, distance and displacement have been distinguished. Displacement occurs in a definite direction which has both magnitude and direction. When the body changes its position in a definite direction then it is called displacement. Besides, distance occurs in any direction.

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