Distance And Displacement

What is Distance?

Distance is the total movement of an object without any regard to direction. We can define distance as how much ground an object has covered despite its starting or ending point.

Let’s understand with the following diagram,

Distance And Displacement

Distance here will be = 4m +3m +5m = 12 m

Distance Formula

\(\Delta d =d_{1} + d_{2}\)

What is Displacement?

It is defined as the change in position of an object. It is of vector quantity and has a direction and magnitude. It is represented as an arrow that points from the starting position to the final position. For Example- If an object moves from A position to B, then the object’s position changes. This change in position of an object is known as Displacement.

Distance And Displacement

Displacement = \(\Delta x = x_{f}- x_{0}\)

\(x_{f}\) = Final Position

\(x_{0}\) = Initial Position

\(\Delta x\) = Displacement

Examples of Distance And Displacement

Question 1.  John to visit Mary in Sydney. She travels 250 miles to North but then back-tracks to South for 105 miles to pick up a friend. What is John’s total displacement?

Answer: John’s starting position  Xi= 0.

Her final position Xf is the distance traveled N minus the distance South.

Calculating displacement, i.e.D.

D = ΔX = (Xf – Xi)

D = (250 mi N – 105 mi S) – 0

D = 145 mi N

Question 2. Julia throw the ball 30 feet North for your dog. Her dog catches the ball and takes it past Julia to her brother, who is standing 5 feet to the South of where she is. Calculate the total displacement of the ball?

Answer: Julia’s initial position with the ball ( Xi) = 0.

As displacement is a vector quantity, ‘direction’ is considered. Xf = (30 feet S – 25 feet N),

so, Xf = -5 feet S.

D = ΔX = (Xf – Xi)

D = (30 ft S – 25 ft S) – 0

D = 5 ft S of julia’s starting position

Practise This Question

The near point of a hypermetropic eye is 1 metre, what is the power of the lens required to correct this defect? (Assuming that the near point of a normal eye as 25cm)