- Relationship between relative positions and velocities
- This is how a child sees it from a moving car
- Relative speed between the motorcycle and the car
- -Exercise resolved
- Exercise 1
- Solution
- References
The relative speed of an object is that which is measured with respect to a given observer, since another observer can obtain a different measurement. Velocity always depends on the observer who measures it.
Therefore, the speed of an object measured by a certain person will be the relative speed with respect to it. Another observer can obtain a different value for the speed, even if it is the same object.
Figure 1. Scheme representing point P in motion, seen from the reference systems A and B. Source: own elaboration.
Since two observers A and B moving relative to each other may have different measurements of a third moving object P, it is necessary to look for a relationship between the positions and velocities of P seen by A and B.
Figure 1 shows two observers A and B with their respective reference systems, from which they measure the position and speed of object P.
Each observer A and B measures the position and velocity of the object P at a given instant of time t. In classical (or Galilean) relativity the time for observer A is the same as for observer B regardless of their relative speeds.
This article is about classical relativity that is valid and applicable to most everyday situations in which objects have speeds much slower than that of light.
We denote the position of observer B with respect to A as r BA. Since position is a vector quantity we use bold to indicate it. The position of the object P with respect to A is denoted as r PA and that of the same object P with respect to B r PB.
Relationship between relative positions and velocities
There is a vector relationship between these three positions that can be deduced from the representation in figure 1:
r PA = r PB + r BA
If we take the derivative of the previous expression with respect to time t we will obtain the relationship between the relative velocities of each observer:
V PA = V PB + V BA
In the previous expression we have the relative speed of P with respect to A as a function of the relative speed of P with respect to B and the relative speed of B with respect to A.
Similarly, the relative velocity of P relative to B can be written as a function of the relative velocity of P relative to A and the relative velocity of A relative to B.
V PB = V PA + V AB
It should be noted that the relative velocity of A with respect to B is equal to and contrary to that of B with respect to A:
V AB = - V BA
This is how a child sees it from a moving car
A car goes on a straight road, which goes from west to east, with a speed of 80 km / h while in the opposite direction (and from the other lane) a motorcycle is coming with a speed of 100 km / h.
In the back seat of the car is a child who wants to know the relative speed of a motorcycle that is approaching him. To find out the answer, the child will apply the relationships that he has just read in the previous section, identifying each coordinate system in the following way:
-A is the coordinate system of an observer on the road and the speeds of each vehicle have been measured with respect to it.
-B is the car and P is the motorcycle.
If you want to calculate the speed of motorcycle P with respect to car B, the following relationship will be applied:
V PB = V PA + V AB = V PA - V BA
Taking the west-east direction as positive we have:
V PB = (-100 km / h - 80 km / h) i = -180 km / h i
This result is interpreted as follows: the motorcycle is moving relative to the car with a speed of 180 km / h and in the - i direction, that is, from east to west.
Relative speed between the motorcycle and the car
The motorcycle and the car have crossed each other following their lane. The child in the back seat of the car sees the motorcycle moving away and now wants to know how fast it is moving away from him, assuming that both the motorcycle and the car maintain the same speeds as before crossing.
To know the answer, the child applies the same relationship that was used previously:
V PB = V PA + V AB = V PA - V BA
V PB = -100 km / h i - 80 km / h i = -180 km / h i
And now the bike is moving away from the car with the same relative speed with which it was approaching before they crossed.
The same motorcycle from part 2 is returned maintaining its same speed of 100 km / h but changing its direction. In other words, the car (which continues at a speed of 80 km / h) and the motorcycle are both moving in a positive east-west direction.
At a certain point, the motorcycle passes the car, and the child in the back seat of the car wants to know the relative speed of the motorcycle with respect to him when he sees it passing by.
To obtain the answer, the child applies the relations of relative motion again:
V PB = V PA + V AB = V PA - V BA
V PB = +100 km / h i - 80 km / h i = 20 km / h i
The child from the back seat watches the motorcycle overtaking the car at a speed of 20 km / h.
-Exercise resolved
Exercise 1
A motor boat crosses a river that is 600 m wide and flows from north to south. The speed of the river is 3 m / s. The speed of the boat relative to the river water is 4 m / s to the east.
(i) Find the speed of the boat relative to the river bank.
(ii) Indicate the speed and direction of the boat relative to land.
(iii) Calculate the crossover time.
(iv) How much it will have moved south from the starting point.
Solution
Figure 2. Boat crossing the river (Exercise 1). Source: self made.
There are two reference systems: the solidary reference system on the river bank that we will call 1 and the reference system 2, which is an observer floating on the river water. The object of study is boat B.
The speed of the boat relative to the river is written in vector form as follows:
V B2 = 4 i m / s
The speed of observer 2 (raft on the river) with respect to observer 1 (on land):
V 21 = -3 j m / s
We want to find the speed of the boat relative to land V B1.
V B1 = V B2 + V 21
Answer i
V B1 = (4 i - 3 j) m / s
The speed of the boat will be the modulus of the previous speed:
- V B1 - = (42 + (-3) 2) ½ = 5 m / s
Answer ii
And the address will be:
θ = arctan (-¾) = -36.87º
Answer iii
The boat crossing time is the ratio of the width of the river to the x component of the boat's speed with respect to land.
t = (600m) / (4 m / s) = 150 s
Answer iv
To calculate the drift that the boat had to the south, multiply the y component of the boat's speed with respect to land by the crossing time:
d = -3 j m / s * 150 s = -450 j m
The displacement towards the south with respect to the starting point is 450m.
References
- Giancoli, D. Physics. Principles with Applications. 6th Edition. Prentice Hall. 80-90
- Resnick, R. (1999). Physical. Volume 1. Third edition in Spanish. Mexico. Compañía Editorial Continental SA de CV 100-120.
- Serway, R., Jewett, J. (2008). Physics for Science and Engineering. Volume 1. 7th. Edition. Mexico. Cengage Learning Editors. 95-100.
- Wikipedia. Relative speed. Recovered from: wikipedia.com
- Wikipedia. Relative velocity method. Recovered from: wikipedia.com