Solving Kinematics Problem: Total Distance in Last Sec vs. 3 Secs

[konichiwa2x]

A stone falls freely from rest and the total distance covered by it in the last secon of its motion equals the distance covered by it in the fist 3 seconds of its motion.How long does the stone remain in air?

I got the answer as 3 secs but my book says it is 5 (much more logical too). Can someone help me please?

 

[civil_dude]

Show your work, and then we can tell you where you went wrong.

 

[gunblaze]

Erm.. 3 sec is definitely not the correct ans. I don't think 5 sec sounds logical to me too. Well, wat level of education are you in? I tot it was quite a simple qn.<assuming there isn't any mistake in the qn u ask> Wat i did was 3+1=4 sec? My ans is 4 and i tot this was nothin to do wit kinematics.

 

[lightgrav]

gun, the problem never said that there was zero fall time
in between the first 3 seconds and the last 1 second ...
I got to class 3 seconds late today, and I closed my book during the last second of it. But there was close to an hour of class between them.

how fast is it going at the end of the first 3 seconds?

 

[tim_lou]

the first 3 seconds it covers a distance of $$.5g3^2$$
equate that to:
$$.5gt^2-.5g(t-1)^2$$
solve it, got
t=5

 

[gunblaze]

gun, the problem never said that there was zero fall time
in between the first 3 seconds and the last 1 second ...
I got to class 3 seconds late today, and I closed my book during the last second of it. But there was close to an hour of class between them.

how fast is it going at the end of the first 3 seconds?

Ooops.. Guess i misinterprete the qn.

 

[Farsight]

konichi: You can get the "feel" for this by telling yourself the stone accelerates at around 10 meters per second.

In the first second its average speed is 5m/s, so it travels 5 meters.
In the second second it starts at 10m/s and ends at 20m/s so its average speed is 15m/s, so it covers 15m. In the third second it covers 25m. So in the first three seconds it covers 45m.
Moving on, it covers 35m in the fourth second, and... 45m in the fifth second. Aha!

Now you can apply the equations (see Tim Lou's post) and know whether the answer you come up with is in the right ball park.

Note that the .5 and the g cancel out because the rate of acceleration is academic, and what this question really is asking is "What number t squared, minus t-1 squared, equals 3 squared". It's

t² - (t-1)² = 9, which works out to
t² - (t² -2t + 1) = 9 which reduces to
2t - 1 = 9.

 

[konichiwa2x]

ok got it now! thanks everyone