A runner covers one lap of a circular track 40.0 m in diameter in 62.5 s. Suppose initially the runner starts by going in the north direction.

Assume the speed is constant for the first half-lap and then constant for the 2nd half-lap. Suppose instead the runner ran these same speeds and distances in a straight line (rather than turning around the track). Sketch the position vs. time graph for this situation.

Well, let's put our running shoes on and have some fun with this question!

Since the runner starts by going in the north direction, let's divide the time into two halves - the first half-lap and the second half-lap.

In the first half-lap, the runner covers half the distance of the circular track. So, the distance covered is 1/2 * 40.0 m = 20.0 m. Since the time taken is 62.5 s, the speed can be calculated as distance divided by time, which gives us:

Speed = 20.0 m / 62.5 s ≈ 0.32 m/s

Now, in the second half-lap, the runner covers the remaining half of the distance of the circular track. Again, the distance covered is 20.0 m, but this time, the time taken will also be 62.5 s. So, the speed remains the same:

Speed = 20.0 m / 62.5 s ≈ 0.32 m/s

If the runner were to run these same speeds and distances in a straight line, the position vs. time graph would look like a straight line with a constant slope.

But because we're talking about a circular track, the position vs. time graph will be a bit more interesting! It will look like a sinusoidal wave because of the continuous circular motion the runner is making.

Picture a wave, but instead of the usual peaks and troughs, you'll have crests and valleys representing the laps and half-laps respectively. The graph will start at the bottom (representing the runner starting north), and then form a crest, a valley, and so on until the end of the time interval.

So, grab a pen and paper, and let your inner artist create a beautiful representation of the runner's position vs. time graph on that circular track. Embrace the curves, and let humor be your guide as you sketch it out!

To sketch the position vs. time graph for the runner who runs at a constant speed in a straight line, we need to understand the distances covered in each half-lap and the corresponding times.

First, let's find the distances covered in each half-lap if the diameter of the circular track is 40.0 m. Since the diameter is 40.0 m, the circumference of the circle is given by C = πd = 3.14 × 40.0 = 125.6 m.

In the first half-lap, the runner covers half the circumference of the circle, which is 0.5 × 125.6 = 62.8 m.

In the second half-lap, the runner covers the remaining half of the circumference, which is also 0.5 × 125.6 = 62.8 m.

Now that we know the distances covered in each half-lap, we can calculate the times taken to cover those distances. Let's assume the runner covers each half-lap at a constant speed and take the total time of 62.5 s as the sum of the two half-lap times.

Let's call the time taken for the first half-lap t_1 and the time taken for the second half-lap t_2. Since the speeds are constant, we can use the formula time = distance / speed to find the times.

For the first half-lap:
t_1 = 62.8 m / speed
For the second half-lap:
t_2 = 62.8 m / speed

The total time is given as 62.5 s, so we have:
t_1 + t_2 = 62.8 m / speed + 62.8 m / speed = 62.5 s

Now, let's sketch the position vs. time graph. Since the runner is running at a constant speed, the position vs. time graph will be a straight line with a positive slope.

On the x-axis, we have the time taken (t) and on the y-axis, we have the position (x) of the runner.

At t = 0, the runner starts at position x = 0.

For the first half-lap, the runner covers a distance of 62.8 m in a certain time t_1.

Therefore, at time t_1, the runner will be at position x = 62.8 m.

After that, for the second half-lap, the runner covers the same distance of 62.8 m in a certain time t_2.

Therefore, at time t_1 + t_2 = 62.5 s, the runner will be at position x = 2 * 62.8 m = 125.6 m.

So, the position vs. time graph will be a linear graph starting at (0, 0) and ending at (62.5 s, 125.6 m).