Question

I'm a bit confused about this kinematics challenge question:

As a space shuttle burns up its fuel after take-off, it gets lighter and lighter and its
acceleration larger and larger. Between the moment it takes off and the time at which it
has consumed nearly all of its fuel, is the magnitude of the average velocity larger than,
equal to, or smaller than half its final speed?

But isn't the magnitude of a velocity vector the same thing as speed since you removed direction??? I'm confused.

Answer 1

The problem might have asked if the average speed is equal to, larger, or smaller than the final speed, without actually changing the substance of the question. Or, it might have compared average velocity to the final velocity. As it were, it compared average velocity to final speed.

Answer 2

Thanks for the quick reply. I'm still a bit confused though, but getting there.

So is the question asking us to compare the value of the average velocity vector (a straight line) with the value of the actual speed (a curved line due to non-constant acceleration) at each point in time? This would make sense.

I think what is throwing me is the use of the term "final speed" which implies the speed when \[ t = t_{final}\]
It doesn't make sense to me to compare the variable velocity at any time t with the constant value of the speed at t_final.

Answer 3

You are right that velocity is a variable. But so is speed. Speed (in one dimension) is just the absolute value of the velocity. (In two or three dimensions we must use the pythagorean theorem to calculate the magnitude, but that is a two or three dimensional form of absolute value) I think the problem is that speed and velocity seem to be slightly misunderstood. Let me see if I can devise an example to be sure you have it correct.

Suppose a car traveling in a straight line due east has a constant speed of 50mph. What is the velocity? Velocity is a constant <50mph, due east>. (In rectangular form: <50mph, 0mph>)

Suppose a car is at rest and increases it speed while traveling in a straight line due east with constant acceleration. At some point during this acceleration, we measure the speed. Let that speed be 35mph. The velocity at that point is <35mph, due east>. Let us measure the speed again a moment later and see that the speed is 40mph. The velocity is <40mph, due east>. (In rectangular form: <40mph, 0mph>)

As you can see, velocity and speed are both variables, even though in the first case they hold a constant value and in the second case they increase at a constant rate.

Last example. Suppose a car is traveling north west with constant speed of 25mph. Let the eastward direction be the positive x axis and the northward direction be the positive y axis. The velocity vector (in polar coordinates, which I have used in the previous examples) is <25mph, north west>. The x component of the velocity is smaller than than 25mph, and so is the y component. Therefore, if we state the velocity as a rectangular vector we get this:\[<\frac{ 25 }{ \sqrt{2} },\frac{ 25 }{ \sqrt{2} }>.\]

Answer 4

Thanks so much for your help Wayne :). Through a combination of your help, and re-reading the problem and its solution...I think I finally understand this problem...and it actually is worded pretty weird IMO.

What they are actually asking is for us to compare the average velocity of the constant acceleration case (straight line) at every conceivable point point in time between t=0 and t=t_final, excluding t_final, with the average velocity of the non-constant acceleration case (x^2 style curved line).

And when it's worded this way...the answer is clear, the average velocity for the constant acceleration case will always be greater until t_final, you can just draw this and see it:



I still don't get how one is supposed to interpret the question like that as it's written though...

Answer 5

Hm, have you taken calculus yet? In the problem, there is only one acceleration curve (straight lines are curves). So the straight line you draw is not to be found in the problem. That makes me think you're missing the calculus view point of view of the average value of a function. For instance, x^2=y. If we look at values for that function from x=0 to x=4, we see that y ranges from 0 to 16. The average is not a straight line from (0,0) to (4,16). The average is a point somewhere on the x^2 curve. To find the average, we need to integrate the function, and we find that there are a lot more higher value y axis values between the x axis range of 2 to 4 than between 0 to 2.