Question

Suppose your average speed is 22 m/s for the first 130 minutes of a 3 hour trip. If you want your average speed for the whole trip to be 18 m/s, what must be the average speed for the remaining 50 minutes of your trip?

Answer 1

Question #54289, Physics / Mechanics | Kinematics | Dynamics |

Suppose your average speed is $22\,\mathrm{m/s}$ for the first 130 minutes of a 3 hour trip. If you want your average speed for the whole trip to be $18\,\mathrm{m/s}$, what must be the average speed for the remaining 50 minutes of your trip?

Answer:

The average speed for 3 hours is defined:

$$v = \frac{v_1 t_1 + v_2 t_2}{t}$$

where $v_1 t_1$ – the distance passed during the first part of trip and $v_2 t_2$ – the distance passed during the second part of trip, $t$ – the total time.

After substitution of all known parameters the equation becomes as follow:

$$18\,\mathrm{m\,s^{-1}} = (22\,\mathrm{m\,s^{-1}} \times 130\,\mathrm{min} + v_2 \times 50\,\mathrm{min}) / 180\,\mathrm{min}$$

Then,

$$\frac{[(3240 - 2860)/50]\,\mathrm{m\,s^{-1}} = v_2}{v_2 = 7.6\,\mathrm{m\,s^{-1}}}$$

Thus, be the average speed for the remaining 50 minutes should be $7.6\,\mathrm{m\,s^{-1}}$.

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