First slide

Rectilinear Motion

Question

A particle moving in a straight line covers half the distance with speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7 .5 m/s respectively. The average speed of the particle during this motion is

Difficult

A

4.0 m/s
B

5.0 m/s
C

5.5 m/s
D

4.8 m/s

Solution

If t₂ and 2t₂ are the time taken by particle to cover first and second half distance, respectively, then
$\begin{aligned} \begin{aligned} t_{1} = \frac{x / 2}{3} = \frac{x}{6} . . . . . . (i) \\ x_{1} = 4.5 t_{2} and x_{2} = 7.5 t_{2} \\ So, x_{1} = x_{2} = \frac{x}{2} \Rightarrow 4.5 t_{2} + 7.5 t_{2} = \frac{x}{2} \\ t_{2} = - \frac{x}{24} . . . . . . . (i i) \\ Total time t_{1} + x_{2} + 2 t_{2} = \frac{x}{6} + \frac{x}{12} = \frac{x}{4} \\ A L T E R N A T E M E T H O D : \\ A V E R A G E S P E E D = \frac{T O T A L D I S T A N C E T R A V E L L E D}{T O T A L T I M E O F T R A V E L} \\ = \frac{X}{\frac{X / 2}{V_{1}} + \frac{X}{V_{2} + V_{3}}} = \frac{2 V_{1} (V_{2} + V_{3})}{2 V_{1} + V_{2} + V_{3}} \\ = 4 m s^{- 1} \end{aligned} \end{aligned}$
$So, average speed 4 m / \sec .$

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