Over 80% Couldn't Answer This Question Correctly!
ฝัง
- เผยแพร่เมื่อ 28 ส.ค. 2024
- On a small planet with no atmosphere, an object falls 5m in the first second.
How far does it travel in the next second?
Thanks for watching,
Lewis
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These types of shorts are really good, can we get more?
Yes!
How would you visualise a graph like this so quickly. How do you know it is exact 3 triangles?
s=5, u=0, t=1, a=?, so in s=ut + 1/2at^2, 5=0(1) + 1/2a(1)^2, so a=2(5)/1=10ms^-2
at t=2, s=1/2(10)(2)^2, s=1/2(40)=20m,
so in 2nd second distance (s) is 20-5=15m
although on University Challenge you only have a few seconds to answer and cant write anything, so would prob have to visualise the v-t graph in your head...
Thanks!
nice explanation the pronunciation also hearable
For those curious, you could also do this with linear algebra and calculus /s
Holy moly this is a next level of smart! 🧠 Thank you!
I literally just saw this poll lol
Lol same.
5m/s^2 so after one second: 5m x 1^2 = 5m. After two seconds: 5m x 2^2 = 5m x 4 = 20m total in two seconds. Subtract the first 5m for the first seconds and you're at 15m. Now knowing TH-cam comments, I've just made a complete fool of myself... :)
Not a fool at all. A very sensible way of working through a correct solution. But I thought for this I'd visualise it using the v-t graph to make it easier to explain.
are you using s=ut + 1/2 at^2, where u=0?
@@davidsweeney111 I'll be honest... physics isn't my thing. ;) I know that acceleration here is something like 9.8m/s^2 so really I just took the equation and substituted the 9.8 with 5 and ran with it. Based on a Google , I suspect you're using something a more sophisticated but perhaps unneeded formula for a this example; nice to know about though... Thanks
@@Clark-Mills but isnt it only 9.8 on earth? and this cant be earth because it says no atmosphere
@@am-hm4oz Yup, just explaining my reasoning using the Earthly equation reapplied to the posed question.
See original comment. :)
The correct answer is actually 15m-2(initial velocity) , since no initial velocity is given, we can't know what contribution of the velocity was due to acceleration or was already there
Tip:
s=u+(1/2)a(2t-1)
u is the initial velocity
s is the distance travelled in the "(t)th" second
t is the t in "(t)th) second"
[u=0 as the object is falling]
t=1:
5=(1/2)a
a=10
t=2:
s=5(3)
s=15m
How would you do this with suvat tho?
Use your SUVAT equations, in particular S=Ut +(Gt^2 ) (0.5), with t being equal to 2s and U=0 metres per second . Then, subtract 5 metres from your answer.
Your acceleration ought to be 10 metres per second per second. Use 5 metres =Ut +(Gt^2 ) (0.5), with t being equal to 1s and U=0 metres per second to validate it.
Of course, I'm assuming that the object falls from rest.
s=u+(1/2)a(2t-1)
I don't think the question is clear enough. No atmosphere doesn't mean no air resistance. Correct me if I am wrong.
if there’s no atmosphere there are no air particles for the object to collide with, so there’s no air resistance
Apparently most people think the terminal velocity with no atmosphere is 5 m/s.
Yes. If there was no atmosphere then there wouldn't be any drag so it would never reach a terminal velocity.
Great qn