Answer:

\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}†Given:−

Graph shows the positions of a body at different times.

\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}†ToFind:−

Speed of the body,

(i) A to B

(ii) B to C

(iii) C to D

\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}†Solution:−

We know that,

\boxed{\pink{\SF Speed \: = \: \dfrac{Distance}{Time}}}Speed=TimeDistance

First let's take,

A to B

Distance covered from A to B =

Final distance covered - intial start =

3 - 0 = 3cm

Time taken =

Final time taken taken - starting time =

5 - 0 = 5s

Speed =

\sf Speed \: = \: \dfrac{3}{5}Speed=53

Therefore,

\boxed{\purple{\textsf{Speed from A to B = 0.6 cm/s.}}}Speed from A to B = 0.6 cm/s.

Second let's take,

B to C

Distance covered from B to C =

Final distance covered - INITIAL start =

3 - 3 = 0cm

This implies that,

Body is at rest.

Time taken =

Final time taken - starting time =

7 - 5 = 2s

Speed =

Since, distance i.e numerator is 0.

Therefore,

\boxed{\purple{\textsf{Speed from B to C = 0 m/s i.e at rest.}}}Speed from B to C = 0 m/s i.e at rest.

Third let's take,

C to D

Distance covered from A to B =

Final distance covered - intial start =

7 - 3 = 4cm

Time taken =

Final time taken taken - starting time =

9 - 7 = 2s

Speed =

\sf Speed \: = \: \dfrac{4}{2}Speed=24

Therefore,

\boxed{\purple{\textsf{Speed from C to D = 2 cm/s.}}}Speed from C to D = 2 cm/s.

Finally,

Speed from A to B = 0.6 cm/s.

Speed from B to C = 0 cm/s.

Speed from C to D = 2 cm/s.