Physics Help Forum Why total time taken s₁/v₁ + s₂/v₂ + s₃/v₃ in average speed?

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Jul 18th 2018, 09:03 AM #1 Junior Member   Join Date: Jul 2018 Posts: 14 Why total time taken s₁/v₁ + s₂/v₂ + s₃/v₃ in average speed? As we know the Average speed = Total distance / Total time but why in the case below Distance traveled = s₁ + s₂ + s₃ and total time taken = s₁/v₁ + s₂/v₂ + s₃/v₃ How is it possible? Could you explain please? so, altogether Vav = s₁ + s₂ + s₃ / ( s₁/v₁ + s₂/v₂ + s₃/v₃)
 Jul 18th 2018, 09:58 AM #2 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 341 speed = distance / time Therefore, time = distance/speed Let's assign some symbols to these quantities... time = t speed = v (from 'velocity') distance = s (don't ask... people just like using s for distance. it crops up a lot in maths too for path lengths and things like that) The equation above described in symbols is: $\displaystyle t = \frac{s}{v}$ Let's use the numbers 1, 2 and 3 to designate the three parts of a journey. These are called "indices". Time for first part of journey: $\displaystyle t_1 = \frac{s_1}{v_1}$ Time for second part of journey: $\displaystyle t_2 = \frac{s_2}{v_2}$ Time for third part of journey: $\displaystyle t_3 = \frac{s_3}{v_3}$ Total time for journey: $\displaystyle t_{total} = t_1 + t_2 + t_3$ Substitute for individual times, we get: $\displaystyle t_{total} = \frac{s_1}{v_1} + \frac{s_2}{v_2} + \frac{s_3}{v_3}$ The total distance is: $\displaystyle s_{total} = s_1 + s_2 + s_3$ Therefore, the average speed is: $\displaystyle \bar{v} = \frac{s_{total}}{t_{total}} = \frac{s_1 + s_2 + s_3}{\frac{s_1}{v_1} + \frac{s_2}{v_2} + \frac{s_3}{v_3}}$ Indranil likes this.

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