The displacement x versus time graph is shown below.

(A) The average velocity during 0 to 3 s is 10 m/s
(B) The average velocity from 3 to 5 s is 0 m/s
(C) The instantaneous velocity at t = 2 s is 5 m/s
(D) The average velocity during 5 to 7 s and instantaneous velocity at t = 6.5 s are equal
(E) The average velocity from t = 0 to t = 9 s is zero
Choose the correct answer from the options given below:

First, remember:

• Slope of x-t graph = velocity
• Straight line → constant velocity
• Horizontal line → zero velocity
• Average velocity = (final position − initial position) / time

(A) Average velocity from 0 to 3 s

From the graph:

$$At\ t=0,x=0$$

$$At\ t=3,x=5$$

So,

$$v_{avg}=\frac{5-0}{3}=\frac{5}{3}\approx1.67\text{ m/s}$$

Given statement says $$10m/s,$$ which is clearly wrong.

So, (A) is false

(B) Average velocity from 3 to 5 s

From graph:

• $$At\ t=3,x=5$$
• $$At\ t=5,x=5$$

So displacement does not change

$$v_{avg}=\frac{5-5}{5-3}=0$$

This means the object is at rest.

So, (B) is true

(C) Instantaneous velocity at t=2 s

At t=2, the graph lies on the straight line between:

• $$(1,-5)and(3,5)$$

Slope (velocity) of this line:

$$v=\frac{5-(-5)}{3-1}=\frac{10}{2}=5\text{ m/s}$$

Since it’s a straight line, velocity is constant in that region.

So, (C) is true

(D) Compare velocities

Average velocity from 5 to 7:

• $$At\ t=5,x=5$$
• $$At t=7,x=0$$

$$v_{avg}=\frac{0-5}{7-5}=\frac{-5}{2}=-2.5$$

Instantaneous velocity at 6.5 s:

This lies between t=6 and t=7, where graph is a straight line from:

• (6,10)to (7,0)

$$v=\frac{0-10}{7-6}=-10$$

Clearly:

$$-2.5\ne-10$$

So, (D) is false

(E) Average velocity from 0 to 9 s

• At t=0, x=0
• At t=9, x=0

$$v_{avg}=\frac{0-0}{9-0}=0$$

So, (E) is true

Final Answer:

(B), (C), (E) only