Question

To qualify for the finals in a racing event, a race car must achieve an average speed of 245 km/h on a track with a total length of 1,700 m. If a particular car covers the first half of the track at an average speed of 222 km/h, what minimum average speed must it have in the second half of the event in order to qualify?

Answer 1

Answer on Question #45692, Physics, Mechanics | Kinematics | Dynamics

To qualify for the finals in a racing event, a race car must achieve an average speed of 245 km/h on a track with a total length of 1,700 m. If a particular car covers the first half of the track at an average speed of 222 km/h, what minimum average speed must it have in the second half of the event in order to qualify?

Solution:

The average speed during the course of a motion is often computed using the following formula:

$$\text{Average Speed} = \frac{\text{Distance Traveled}}{\text{Time of Travel}}$$ $$v_{av} = \frac{d}{t_1 + t_2}$$ $$v_1 = 222 \text{ km/h},$$ $$v_{av} = 245 \text{ km/h},$$ $$d = 1700 \text{ m},$$ $$v_2 = ?$$ $$\text{time} = \frac{\text{distance traveled}}{\text{speed}}$$ $$t_1 = \frac{d_1}{v_1} = \frac{d}{2v_1}$$ $$t_2 = \frac{d_2}{v_2} = \frac{d}{2v_2}$$

Thus,

$$v_{av} = \frac{d}{t_1 + t_2} = \frac{d}{\frac{d}{2v_1} + \frac{d}{2v_2}} = \frac{1}{\frac{1}{2v_1} + \frac{1}{2v_2}}$$ $$\frac{1}{v_1} + \frac{1}{v_2} = \frac{2}{v_{av}}$$ $$\frac{1}{v_2} = \frac{2}{v_{av}} - \frac{1}{v_1}$$

So,

$$v_2 = \frac{v_1 v_{av}}{2v_1 - v_{av}} = \frac{222 \cdot 245}{2 \cdot 222 - 245} = 273.3 \text{ km/h}$$

Answer: $v_2 = 273.3 \text{ km/h}$

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