Question

a particle covers 60 km at a uniform speed at 60 km/h. what should be it's speed for the next 120 km if the average speed for the entire journey is 60 km/h.

Answer 1

Answer on Question #42326 – Physics – Mechanics | Kinematics | Dynamics

a particle covers $60~\mathrm{km}$ at a uniform speed at $60~\mathrm{km/h}$. what should be it's speed for the next $120~\mathrm{km}$ if the average speed for the entire journey is $60~\mathrm{km/h}$.

Solution:

$S_{1} = 60~\mathrm{km}$ – first part of the distance;

$S_{1} = 120~\mathrm{km}$ – second part of the distance;

$V_{1} = 60\frac{\mathrm{km}}{\mathrm{h}}$ – speed on the first part of the distance;

$V_{2}$ – speed on the second part of the distance;

$V_{a} = 60\frac{\mathrm{km}}{\mathrm{h}}$ – average speed for entire journey;

The average speed is the total distance divided by the total travel time.

$$V_{a} = \frac{S}{t} \quad (1)$$

The total distance is

$$S = S_{1} + S_{2} \quad (2)$$

The total time is

$$t = t_{1} + t_{2} = \frac{S_{1}}{V_{1}} + \frac{S_{2}}{V_{2}} = \frac{S_{1}}{V_{1}} + \frac{S_{2}}{V_{2}} = \frac{S_{1}V_{1} + S_{2}V_{2}}{V_{1}V_{2}}$$

(3) and (2) in (1):

$$V_{a} = \frac{S_{1} + S_{2}}{\frac{S_{1}V_{1} + S_{2}V_{2}}{V_{1}V_{2}}} = \frac{V_{1}V_{2}(S_{1} + S_{2})}{S_{1}V_{1} + S_{2}V_{2}}$$ $$S_{1}V_{1}V_{a} + S_{2}V_{2}V_{a} = V_{1}V_{2}(S_{1} + S_{2})$$ $$V_{2}(V_{1}(S_{1} + S_{2}) - S_{2}V_{a}) = S_{1}V_{1}V_{a}$$ $$V_{2} = \frac{S_{1}V_{1}V_{a}}{V_{1}(S_{1} + S_{2}) - S_{2}V_{a}} = \frac{60~\mathrm{km} \cdot 60~\frac{\mathrm{km}}{\mathrm{h}} \cdot 60~\frac{\mathrm{km}}{\mathrm{h}}}{60~\frac{\mathrm{km}}{\mathrm{h}} (60~\mathrm{km} + 120~\mathrm{km}) - 120~\mathrm{km} \cdot 60~\frac{\mathrm{km}}{\mathrm{h}}} = 60~\frac{\mathrm{km}}{\mathrm{h}}$$

Answer: speed on the next $120~\mathrm{km}$ should be $60~\frac{\mathrm{km}}{\mathrm{h}}$

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