Yes, Btw, if you know the displacement eqn ( y= ut + 1/2 at^2) . This simplifies to: y = 1/2 (a*t^2) Put (t =10, y= 50) we get a =1. y= 1/2 (t^2) for t = 30 y = 1/2 * 900 = 450
Yes, but if you look close to the question, there is written nearest meter. So, our answer should only contain meter. Actual answer is 1736.48 (.48 is cm part, so we have to eliminate it by rounding up as below) 1736.48 -> 1736.5 -> 1737 (add 1 if leftmost digit is greater than and equal to 5)
@@KO-nr2rr i can go to any website and round 1736.48 to the nearest number and they will say 1736. This is a big trap, rounding states if the decimal by itself is greater than .5 then you can round up if not you keep it and remove the decimal thats less than .5
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40:04 for this one you could just regression on desmos using the points (0,0), (10,50) and (20,200) then plug in x
Yes, Btw, if you know the displacement eqn ( y= ut + 1/2 at^2) . This simplifies to:
y = 1/2 (a*t^2)
Put (t =10, y= 50) we get a =1.
y= 1/2 (t^2)
for t = 30
y = 1/2 * 900
= 450
Not all heroes wear caps 🥹 thank you sir for putting all of the lessons here
Thanks you a lot 🎉
10:28 the answer is 1736.
Yes, but if you look close to the question, there is written nearest meter. So, our answer should only contain meter. Actual answer is 1736.48 (.48 is cm part, so we have to eliminate it by rounding up as below)
1736.48 -> 1736.5 -> 1737 (add 1 if leftmost digit is greater than and equal to 5)
@@KO-nr2rr i can go to any website and round 1736.48 to the nearest number and they will say 1736. This is a big trap, rounding states if the decimal by itself is greater than .5 then you can round up if not you keep it and remove the decimal thats less than .5