Lucy is using a programmable treadmill. She inputs a top speed into the treadmil

Question

Lucy is using a programmable treadmill. She inputs a top speed into the treadmill and presses "start." The treadmill accelerates from rest at 0.100 m/s2 until it reaches that speed, after which it stays at that constant speed. Later, when she is getting tired, she hits the "stop" button and the treadmill slows down at a rate of 0.200 m/s2 until it comes to a stop. The treadmill tells her that she traveled the equivalent of 3.00 kilometers in 15.0 minutes while the treadmill was in motion. (a) If Lucy had been running forward in a straight line on the ground (instead of in place on the treadmill), what would be her average velocity in meters per second over the whole 15.0 minutes? (b) What would be her average acceleration over the 15.0 minutes? (c) What was the top speed of the treadmill? (d) How long after pressing "start" did she press "stop?"

Explanation / Answer

How about this: I'll outline my approach and give you the answers I got, OK?

I used graphing on a standard x-y coordinate system, where

y is the velocity of the treadmill and
x is the time

First, I graphed the line

(1) y = 0.100 * x; in other words, the velocity vs. time as long as it is increasing at a uniform rate;

Then, I graphed a line sloping downward and hitting y = 0 at t = 900s

(2) -y = -0.200 * (900 - stop), where

stop represented the time (as yet unknown, this is just a sketch) at which she pushed the "stop" button.

Then I sketched a horizontal line at

(3) y = ytop,

representing the top speed of the treadmill. This line I drew below the point where (1) and (2) intersected, because I think it did run at a constant speed before she hit the "stop" button.

Consider the area bounded by the lines (1), (2), (3), and y = 0. Since the first 3 lines represent velocity, then the area underneath them must represent the total distance covered. This area therefore = 3000m (3km).

You can break this area down into two triangles and one rectangle that representing the 3 phases of her session: Speed going up, speed constant, and speed going down.

The x-axis is time; so let

(4) ttop = the time (in seconds) at which y = ytop the top speed is reached; similarly, let

(5) stop = the time at which she hit the stop button.

Thus ttop and stop represent the t-values at which equations (1) and (2) resp. intersect equation (3).

Using the above variables, we can create equations for the areas of the 3 polygons, which I'll label:

Aup = the area of the triangle formed during "speed going up" phase = 0.5 * ttop * ytop

Aconst = the area of the rectangle formed during the "speed constant" phase = (stop - ttop) * ytop

Adown = the area of the triangle formed during the "speed going down" phase = 0.5 * (900 - stop) * ytop

Therefore, it's given that:

(6) Aup + Aconst + Adown = 3000

Also, since we know the slope values of the acceleration and deceleration, we can express ttop and stop in terms of ytop:

(7) ytop / ttop = 0.100 and

(8) -ytop / (900 - stop) = -0.200

OK!! That's the setup: Then follows the mathematics of substituting the equations for ttop and stop into (6).

You'll get a quadratic equation like this (I'm going to substitute the letter h for ytop):

(9) h^2 - 120 * h + 400 = 0

It has two roots, one which doesn't make any sense and the other is:

(10) h = 3.43, which does make sense.

In other words, ytop = 3.43m/s <<<===Answer to (a)

And from that, you can calculate ttop and stop:

(11) stop = 882.8s <<<===Answer to (b)

[I plugged these values back into (6) to verify that the ytop yielded the desired areas (distances) covered in the 3 phases added up to 3000m.]

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