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monkfishkev
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Homework Statement


An object starts from position X = 0 and moves along a straight line with its position given by x = 4t^2 - t^3 where t is the time taken to reach position x.
Find:
i) The time at which the velocity is zero
ii) The time at which the acceleration is zero
iii) The average velocity over the first three seconds.

2. The attempt at a solution
Please can somebody let me know if these answers are correct
i) 0.53s
ii) 0.27s
iii) -37s

Thanks
 
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Discussion

It's good if you try to sketch the graph of the function where you can evaluate you answers.
 
Hi monkfishkev
1) no, this is wrong, and the velocity is 0 more than once, it is 0 at 0 how are you solving this ?
2) wrong too, but you should write how you solved it instead of giving the numerical result so that we know if your method is wrong or if you just messed the numerical result (I suspect the former)
3) wrong too

Please explain how you understood the problem and how you solved it, the numerical result is much less important than showing how you get it

Cheers...
 
OK I've tried it a second time and come out with a different answer.
I differentiate x = 4t^2 - t^3 to become v = 8t - 3t^2
Making v = 0 I get either v = 2.6 or v = 0. How do I know which one to use.

I then differentiate further to solve for acceleration. a = 8 - 6t
when making a = 0 I get 1.3 seconds

Am I now on the right track?
 
monkfishkev said:
OK I've tried it a second time and come out with a different answer.
I differentiate x = 4t^2 - t^3 to become v = 8t - 3t^2
Making v = 0 I get either v = 2.6 or v = 0. How do I know which one to use.

I then differentiate further to solve for acceleration. a = 8 - 6t
when making a = 0 I get 1.3 seconds

Am I now on the right track?

Well I think the more -interesting- time would be the t=2.66 since the object was at origin and not moving at t = 0
for 3 you can just use mean value theorem for integrals right?
average f(x) from a to b = integral from a to b f(x)dx/b-a
 
Yes you are on the right track
for (1), as madah12 says, the case where t is non 0 is 'more interesting' so to speak. I would just formulate my answer stating that there are two good answers and give both instead of picking one. (if for some reason you absolutely must pick one, than the non zero one looks better yes)
(2) is ok too
for (3) you don't need anything fancy: you have the position at t=0, the position at t=3, the average speed is the distance by the time, so
[(position at 3) - (position at 0)]/3
position at 3 is 0, so your average speed is simply x(3)/3

Cheers...