Back to: PHYSICS SS2

**Welcome to class! **

In today’s class, we will be talking about position, distance and displacement. Enjoy the class!

**Position, Distance and Displacement**

**POSITION**

Position is a point in space, is determined by its distance and direction from other points. The statement of position is accomplished by means of a frame of reference or a point of reference, which is called the Origin. Position is a location of a point using rectangular coordinates, x and y axes. When writing the coordinates of points, x is written first followed by y.

**WORKED EXAMPLE**

(1) Point A (2, -3) has the value of x as 2 and that of y as – 3. Represent this on the Cartesian plane.

**SOLUTION **

**NOTE:**

The x – coordinate is also known as abscissa while the y – coordinate is known as ordinate.

**DISTANCE**

Distance is the product of average speed and time.

Distance = average speed x time

Distance is also defined as the length of space covered by an object when it changes position from one point to another. Note that the direction is not specified hence distance is a scalar quantity. The S.I unit is the metre (m).

**DISPLACEMENT**

Displacement is the distance travelled in a constant direction.

Displacement average velocity x time

Displacement is also defined as the distance a body moves in a specified direction. It is, therefore, a vector quantity.

It is denoted by the letter **‘s’** and has both magnitude and direction. The SI unit for displacement is the metre (m).

**WORKED EXAMPLES**

**(1)** A body moving with a constant velocity along a straight line ABC takes 45 minutes to go from A to B and 10 minutes from B to C. if Ac is 3km find AB

**SOLUTION**

Average Velocity:

But displacement s = average velocity, v x time, t

Therefore: Displacement AB = 3/55 x 45 = 2.45 km.

**(2)** A man walks 1 km due East and then 1 km due North. Find the displacement.

**SOLUTION**

R^{2} = 1^{2} + 1^{2}

R^{2 }= 2

R =

tan = = 1

= 45°

Therefore the displacement R = N 45° E

**(3)** Calculate the distance in metres covered by a body moving with a uniform speed of 180 km/h in 30 seconds.

**SOLUTION**

Given parameters

Time = 30 seconds

Uniform speed = 180km/r

Converting from km/hr to m/s

Therefore

**Distance covered**= Speed x time

=50 x 30

=1500 m

**(4)** A body is uniformly accelerated from rest to a final velocity of 100m/s in 10 seconds. Calculate the distance covered.

**SOLUTION**

Given parameters

Velocity (u) = 100m/s

Time = 10seconds

Note: Velocity is given so we apply the equation of motion

Therefore (a = 10m/s^{2})

S=ut+ ½ at^{2}

Substituting:

=0×10+ ½ ×10×10^{2}

= = 500 m

**Distance between two coordinates points**

Let P P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be any two points. The distance between P and Q is found by completing the right-angled triangle PQR with the right angle at R by applying Pythagoras Theorem.

**WORKED EXAMPLES**

**(1)** Find the distance between the two coordinates A(3, 2) and (6, 4)

**SOLUTION**

The distance (d) between two points (x_{2} – x_{1}) and (y_{2 }– y_{1}) is defined by:

**(2)** Find the value of x, given the following coordinates A(1, -2) and B(x, -5) with a distance of 5.

The distance (d) between two points (x_{2} – x_{1}) and (y_{2 }– y_{1}) is defined by:

Given parameters:

d = 5

(x_{1} – y_{1}) = (1, -2) and (x_{2 }– y_{2}) = (x, -5)

Substituting:

Collect like terms

25 – 9 =

16 =

Take square root of both sides

x = 5

Therefore x = 5

A(1, -2) and B(5, -5)

In our next class, we will be talking about** Physical Quantities.** We hope you enjoyed the class.

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