A uniformly accelerated car passes three equally spaced traffic signs. The signs

Question

A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 20 m. The car passes the first sign at t = 1.5 s, the second sign at t = 3.9 s, and the third sign at t = 5.2 s.

(a) What is the magnitude of the average velocity of the car during the time that it is moving between the first two signs?

(b) What is the magnitude of the average velocity of the car during the time that it is moving between the second and third signs?

(c) What is the magnitude of the acceleration of the car?

Explanation / Answer

Remember, average velocity is the distance you went divided by the time it took you to cover that distance. All of the signs are spaced 20m apart, remember.

a) What's the average speed between signs #1 and #2? Well, how much distance is there between them, and how much time did it take to go from one to the other? The distance is 20m. The *time* is the time it passed the 2nd sign (what we call "final" t) minus the time it passed the 1st sign (what we call "initial" t). This is 3.7s minus 1.5s, or 2.2s.

So, average speed is 20m / 2.2s = 9.1m/s

b) What's the average speed between #2 and #3? Again, how much distance is there between them, and how much time did it take to go from one to the other? The distance, like last time, is 20m. This time, however, "final" t is 4.9s and "initial" t is 3.7s. So, the duration of time that the car was between the signs is 4.9s - 3.7s, or 1.2s.

So, average speed is 20m / 1.2s, or 16.7m/s

c) This one is a little tough. Acceleration is the change in velocity divided by the time it took for the velocity to change that much. The problem is, we don't know what the actual *velocities* were at the moments that we know *times* for (ie, when the car was passing the signs). We also don't know the *times* or *distances* the car was at the moments when it was traveling at either of those average velocities that we calculated in parts a) and b). Lastly, we don't know if the car had an initial distance or initial velocity, which makes using the old X = 1/2at^2 + V0t + X0 equation problematic.

It turns out that you *can* solve this, but the math is a little complicated. It turns out that, if you know the average speed between points 1 and 2 (which we'll call V12) and the average speed between points 2 and 3 (call it V23), and you know the times that the car passed points 1, 2, and 3 (which we'll call t1, t2, and t3)... and we know all of those things, then average acceleration is given by:

a = 2*(V12 - V23) / [ (t2^2 - t1^2)/(t2 - t1) - (t3^2 - t2^2)/(t3 - t2) ]

Which all comes out to a = 4.5m/s^2

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