Question

An ant crawls in a straight line at a constant speed of 0.24 m/s for a distance of 3.0 m, beginning in the corner of a square classroom. It then turns exactly 90 degrees to the right, and proceeds an additional 4.0 m, reaching the far corner of the same wall from which it began. If the second leg of the journey was crawled in half the amount of time as the first, what was the ant's average speed for the whole trip?

Answer 1

Answer:

vavg = 0.37 m/s

Explanation:

• The average speed is just the relationship between the total distance traveled, and the total time required for that travel , as follows:

       $v_{avg} = \frac{\Delta x}{\Delta t} (1)$

• We know that for the first leg of the journey, the ant crawls at a constant speed of 0.24 m/s, moving 3.0 m.
• We can find the time required for this part, just applying the definition of average velocity, and solving for the time t (which we will call t₁), as follows:

       $t_{1} =\frac{x_{1}}{v_{1} } = \frac{3.0m}{0.24m/s} = 12.5 s (2)$

• From the givens, we know that the time for the second part is exactly the half of the value found in (2), so we can write the total time Δt as follows:

       $\Delta t = t_{1} + \frac{t_{1} }{2} = 12.5 s + 6.25 s = 18.75 s (3)$

• We also know that in the second leg of the journey, the ant traveled 4.0 m, which adds to the 3.0 m of the first part, making a total distance of 7.0 m.
• Per definition of average speed, we can write the following expression as in (1) replacing Δx and Δt by their values, as follows:

       $v_{avg} = \frac{\Delta x}{\Delta t} = \frac{7.0m}{18.75m} = 0.37 m/s (4)$