Question

A particle moving in a straight line covers half the distance with speed of 12m/s . The other half of the distance is covered in two equal time intervals with speed of 7. 5 and 4. 5 . The average speed of the same of the particle during this motion is

Answer 1

Answer:

8m/s

Explanation:

Given :

A particle moving in a straight line covers half the distance with speed of 12m/s . The other half of the distance is covered in two equal time intervals with speed of 7. 5 m/s and 4. 5 m/s.

Solution :

Average speed = $\frac{Total \ distance \ traveled}{Total \ time \ taken}$

Let the total distance be d,

Let the total time be $t_1+t_2+t_3$

⇒ Time taken by the particle to travel half the distance with the velocity 12 m/s = $\frac{distance \ traveled}{speed}$

⇒ $t_1 = \frac{(\frac{d}{2})}{12} = \frac{d}{24}$

⇒ Time taken by the particle to travel half the distance with the velocity 4.5 m/s & 7.5 m/s($t_2=t_3$)= $Distance = speed \times time$

= $\frac{d}{2}=4.5 \times t_2 + 7.5 \times t_3 = 4.5 \times t_2 + 7.5\times t_2 = 12 \times t_2$

⇒ Time $t_2 = \frac{\frac{d}{2}}{12}=\frac{d}{24}$

⇒ Time $t_3 = \frac{\frac{d}{2}}{12}=\frac{d}{24}$

Average speed = $\frac{Total \ distance}{Total \ time}$

⇒ $\frac{d}{t_1+t_2+t_3} = \frac{d}{(\frac{d}{24})+(\frac{d}{24})+(\frac{d}{24})} = \frac{d}{d(\frac{1}{24}+\frac{1}{24}+\frac{1}{24})} = \frac{1}{(\frac{3}{24})} = \frac{24}{3} = 8$ m/s

Answer 2

Answer:

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