1
$\begingroup$

A brick is dropped from the roof of a tall building.After it has been falling for a few seconds ,it falls $40.0$ meters in a $1.00$-s time interval.What distance will it fall during the next $1.00$ seconds? Ignore air resistance.

I've seen on yahoo answer that the solution provided for second part of the question,namely the distance covered in the second interval of $1.00$ second,can be found just by realizing that the midpoint of this interval occurs $1$ second later the midpoint of the first interval(so this would allow us to calculate instantaneous velocity).

However I am a bit skeptical ,because after the midpoint of the first interval the brick's velocity kept raising in magnitude,so I would think that it hasn't necessarily have to be at the midpoint of the second interval exactly $1$ second later after the first one.

This way I would be assuming that the brick's velocity is constant over that interval,while it isn't.

Another point of confusion is that I can find many intervals of $1$ second where the brick falled a distance of $40$ meters..

Question

Can you guys make this clear ?Am I right or wrong ?If I am right how would I solve the problem ?

|cite|improve this question
$\endgroup$
  • $\begingroup$ I think your doubt in the answer referred to is the assertion "When acceleration is constant, the instantaneous velocity at the midpoint of a time period is equal to the average velocity for [the] time period." That is incorrect, but at whatever point in a one second period the average velocity occurs, it will occur exactly one second later in the next such period, so the answer obtained was still correct ! $\endgroup$ – true blue anil Dec 28 '15 at 12:03
1
$\begingroup$

By the well-known relation, the space traversed by the brick follows

$$h(t)=\frac{gt^2}2.$$

You are given that

$$h(T+1)-h(T)=H=\frac{g(2T+1)}2.$$

Then,

$$h(T+2)-h(T+1)=\frac{g(2T+3)}2=H+g.$$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ why is it $2T+1$ shouldn't it be just $1$ ? $\endgroup$ – Mr. Y Dec 28 '15 at 8:59
  • $\begingroup$ There is a seeming anomaly in the equation as the terms are no homogeneous. This disappears if you write $\delta t$ instead of $1$, and the answer is $H+g\,\delta t^2$. $\endgroup$ – Yves Daoust Dec 28 '15 at 9:00
  • $\begingroup$ @Mr.Y: please do the complete substitution by yourself. $\endgroup$ – Yves Daoust Dec 28 '15 at 9:01
  • $\begingroup$ I am not asking for that @Yves Daous.I am asking the concept behind that,because $2T+1$ seems to me as the $1$ second later the two $1$ seconds interval... $\endgroup$ – Mr. Y Dec 28 '15 at 9:02
  • 1
    $\begingroup$ My apologies.Thanks for the clarification.Mathematical arguments are indisputable :) $\endgroup$ – Mr. Y Dec 28 '15 at 9:11
1
$\begingroup$

More simply, the SUVAT equation $v = u + at$ shows that the "final" velocity increases by $a\; (= g)$ every second, thus so must the average velocity.(**)

Hence distance travelled in the next second $ = (40 +g)\cdot1 = 40+g$

(**) Explanation :

Velocities at time, $0,1,2$ seconds: $u, u+g, u+2g$
Average velocity $0-1: u + 0.5g = 40 =$ distance travelled in first second
Average velocity $1-2: u + 1.5g =40 +g =$ distance travelled in next second

|cite|improve this answer
$\endgroup$
  • $\begingroup$ "so must the average velocity": you should prove that. $\endgroup$ – Yves Daoust Dec 28 '15 at 10:08
  • $\begingroup$ Well, I presumed that tackling such a question, OP would easily be able to figure out that in a straight line relationship, for any two points $1$ second apart, the difference in velocity would be the same. $\endgroup$ – true blue anil Dec 28 '15 at 10:23
  • $\begingroup$ The OP was stuck on the question, wasn't he ? $\endgroup$ – Yves Daoust Dec 28 '15 at 10:31
  • $\begingroup$ Ok, I am editing. $\endgroup$ – true blue anil Dec 28 '15 at 10:43
  • $\begingroup$ When you say "velocities at time $0,1,2$ " seconds do you mean velocities during those time intervals ?Or are we assuming that the bricks falls $40$ meters $1$ second later it was released from the building(quite unrealistic ) ? $\endgroup$ – Mr. Y Dec 28 '15 at 12:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.