Kinematics

[Tetra]

Just a (simple) problem that I couln't solve.

 

A motorcycle starts from rest and, moving with uniform accelerations, moves 66m in the sixth second (from time equals 5.0s to time equals 6.0s). What is its acceleration, and what distance does it travel in the 7th second?

 

I'm not really sure how to do this. Why can't we do the following?

So to find acceleration, the equation is

a = Δv/ t, or (v2 - v1) / t.

And v2 = d/t.

So v2 = 66 /1 = 66

So, to find acceleration:

a = (v2 - v1)/t

= (66 - 0) / 6

= 11

 

But that's not right! And a friend told me it was supposed to be 66 / 5.5, but he doesn't explain why either! So I don't get it. Help?

[swansont]

v2 ≠ d/t

 

d/t gives you the average speed

[Tetra]

That's true...

 

Oh I just thought of something! Not sure if it's right, but...

 

If d/ t is average velocity, then, if say we were to draw it on a v/ t graph, the time of 5.5 should have velocity of 66 (because 5.5 if between 5 and 6, then the center point should have the average velocity?)

 

So from that we can calculate acceleration. But I'm not sure about the next part of the question, which is to find how much displacement was covered in the sixth second. I was thinking we could us the equation:

 

d = vt + 1/2 (a) (t)^2

 

But I'm not sure what to put in for time. Should it be one second, because one second elapsed in the 6th second?

[cypress]

Your friend may well be correct, but knowing the answer isn't going to help you understand the concept is it?

 

Though you have stumbled onto something by noting your friends answer is the average of the speed between 5 and 6 seconds divided by the average of the time between 5 and 6 seconds, but my question to you is why did this method work for this problem? Will this method work in the general case, and if not, what cases will this work for?

 

The equation you proposed is the general case and it is correct that you can/should use it. Think more about what the equation means and see if you can answer your question about what values of t to use. Notice how many unknowns you have compared to the known(s) you have and perhaps you can rearrange the formula to reduce it to solve for a.

[Tetra]

This is just me musing:

 

I think the method could only be used for problems with uniform motion, because if it weren't (a straight line), then we can't simply solve for the slope by using a = v/t. In this case, if averages are used for v and t, we can find average acceleration (which is just acceleration, since this is a straight slope)

 

So, in this case, if we use the averages for both v and t, we can solve for a (which is 66/ 5.5). And, since the acceleration is uniform the whole way, the method can be used backwards to solve for distance in the 7th second.

 

So a = v / t

12 = v/ 6.5

= 78

 

And v = d/ t

78 = d/ 1

d = 78

 

...am I right? Or am I just using a confusing method..?

[cypress]

You got it.

 

The general case goes like this:

 

d(6) - d(5) = (1/2)*a*(6^2 - 5^2) so (66*2)/(36-25) = a or a = 12 then use the same formula to solve for d(7)-d(6)

Edited July 7, 2010 by cypress

[swansont]

That's true...

 

Oh I just thought of something! Not sure if it's right, but...

 

If d/ t is average velocity, then, if say we were to draw it on a v/ t graph, the time of 5.5 should have velocity of 66 (because 5.5 if between 5 and 6, then the center point should have the average velocity?)

 

So from that we can calculate acceleration. But I'm not sure about the next part of the question, which is to find how much displacement was covered in the sixth second. I was thinking we could us the equation:

 

d = vt + 1/2 (a) (t)^2

 

But I'm not sure what to put in for time. Should it be one second, because one second elapsed in the 6th second?

 

t in the equation is the total time. But if you figure out a, you can solve it for t=6 and t=7