Question

A particle moves in a straight line obeying the relation x = t(t-1) where x - displacement in m and t = time in sec. What are the distance covered and displacement during t = 2 sec?

Answer 1

I solved for displacement just by putting t = 2 sec. in the given relation as x is the displacement itself. So, x = 2(2-1) = 2 m.

Answer 2

But, not sure for the distance. @Hero @AravindG @zepdrix @austinL @mathmale

Answer 3

Wish I could help @mathslover :/

Good Luck with finding the answer!

Answer 4

np Shadow. Thanks!

Answer 5

May I assume that you're taking Calculus for at least have some experience with the derivative as a rate of change / velocity?

Answer 6

Well, yes, I know basics of calculus. And yes, I have a lot of experience with differentiation but not a great experience with integration.

Answer 7

Though, as per this question/topic is concerned, my calculus concepts are at right point of my mind Sir.

Answer 8

You're asked to find the "distance covered" over (I'm assuming) the time interval [0,2]. Note that the particle can move both to the left and to the right from its starting point on the x-axis. Although its displacement from t=0 to t=2 can be found by subtracting t(0) from t(2), its total distance traveled cannot. How are you going to find the distance traveled in the (+) direction separately from the distance traveled in the negative direction? You must do so, and then must add those two distances together.

So: differentiate x=t(t-1); set the derivative = to 0 and solve for t (your critical value). Why? What next?

Answer 9

That is what confused me.

If I differentiate x = t(t-1) => \(\dfrac{d(t^2 - t)}{dt} = 2t - 1\)

If I set the derivative = 0 , so, I get 2t - 1 = 0 and t = 1/2

Answer 10

I'm not sure that what does it signify? Does it signify that velocity of the particle will be zero at t= 1/2 ?

Answer 11

Yes. For a clearer picture of what is happening, treat x=1/2 as a "critical value:" graph it on a number line subdivided into intervals [0,1/2) U (1/2,2] Determine for which time interval the velocity is negative and for which it is positive.

How far does the particle move during the time interval [0,1/2)? during the time interval (1/2,2]? Add together the absolute values of those two distances to obtain the total distance traveled.

Answer 12

Unfortunately, I need to get off the 'Net. But I'll be back on later today and could possibly help you again later if need be.

Answer 13

Sure Sir. Thanks a lot, I will try to solve it from here.

Answer 14

May i suggest u go through this video? xD

http://www.youtube.com/watch?v=8Ig-e509uRQ

right at 27:41.. see what he does.. its really great!!.. just go through it once :)

Answer 15

Thanks a lot @Mashy for sharing the link. I will have a look at it.

Answer 16

sure.. and do tell me what answer u get.. ll check.. if its right .. also try drawing

xt, and vt graphs for this.. (to get a better insight ).. hope u see xt would be a parabola, vt would be a straight line and a is constant hence horizontal line!

Answer 17

Yep. I drew the xt graph and I got it as a parabola ... the max. negative value for the distance being at time = 1/2 . And yes, vt graph will be a straight line.

I got 2.5 m as the answer, is it right?

Answer 18

yes ..

at t=0 x =0
at t = 0.5 x = -0.25 hence it traveled 0.25 meters in negative x
at t = 2, x = 2 hence it traveled back 0.25 and then traveled additional 2 meters both in positive x

so together its 2+0.25+0.25 = 2.5 m!

Answer 19

But.. ur question wording "during t = 2 sec? "

it makes me think.. that it needs u to find the distances and displacement DURING the 2nd second.. from t =2s to t = 3s .. why is it worded like that :-/

Answer 20

" t =2s to t = 3s ."

i meant " t =1s to t = 2s ".

Answer 21

Not sure, but the book's answer is also 2.5 , so, I guess it meant 0 to 2 sec.

Answer 22

oh ok ok :)

Answer 23

Again: Please be certain to distinguish between
1) total distance traveled, regardless of direction of travel, and
2) the displacement over the period [0,2].

"the book's answer" ... for what? is 2.5. This problem calls for two separate answers, as described above.

Answer 24

Sir, 2.5 for distance travelled (total)

Answer 25

And sorry for not being specific.