Question

In the diagram shown, the displacement(x) of particles is given as a function of time. The particle A is moving under constant velocity of 9 m/s .The particle B is moving under variable acceleration. From time t + 0 s. to t = 6 s., the average velocity of the particle B will be equal to .

• A. 2.5 m/s
• B. 4 m/s
• C. 9 m/s
• D. None

Answer 1

The correct option is C 9 m/s

This is the displacement $\left(x\right)$ of particle and time graph$(d-t)$

Constant velocity $=9m/s$

For Particle A

For Particle B

This particle under variable acceleration.

From time $t+0stot=6s$

Now the average velocity of $B=V_{B}m/s$

V_{avg}=\cfrac{\int^t_09vdt}{\int^t_0dt}$

Now, $V_B=\frac{54}{6-0}=\frac{9\times 6}{6}=9m/s$