Question

The velocity-time graph of a particle moving along a straight line is as shown in the diagram. The rate of acceleration and retardation is constant and is equal to $\text{5} \, \text{m s}^{- 2}$ . If the average velocity during the motion is $\text{20} \, \text{m s}^{- 1},$ then the maximum velocity of the particle is

• A. $\text{20} \, \text{m s}^{- 1}$
• B. $\text{25} \, \text{m s}^{- 1}$
• C. $\text{30} \, \text{m s}^{- 1}$
• D. $\text{40} \, \text{m s}^{- 1}$

Answer 1

Answer: (B)

Average speed $=20 \, \text{m s}^{- 1}$
$\frac{\text{Total distance}}{\text{Total time}} = 2 0 \Rightarrow \frac{\text{Area of graph}}{2 0 + \text{t}} = 2 0$
$⇒ \, \, \, \frac{1}{2} \left(2 0 + \text{t} + 2 0 - \text{t}\right) \times 5 \text{t} = 2 0 \left(2 0 + \text{t}\right) \, \, ⇒ \, \, \text{t} = 5 \text{ s}$ Maximum velocity $=5t=5\times 5=25 \, \text{m s}^{- 1}$