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Mathematics: Post your doubts here!

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You can not use that. The equations of motion are only applicable when the magnitude of acceleration is constant.

For 5ii) We know that particle Q was thrown 2 seconds later. Also, we will use s = ut + 1/2 a t^2 because it is asking about the time when one particle is 5 m higher than the other. The equation gives us both time and displacement, and is according to the question.

Now, when P is 5m higher than Q, we know that displacement of P - displacement of Q should equal 5. (Get used to this concept as there are a lot of such questions).
Using the equation
s = 17(t+2) - 5(t+2)^2
We use t + 2 because P was thrown two second earlier. Thus, it has been travelling two seconds more than Q. Now the equation for Q would be
s = 7t - 5t^2

Now we will eliminate s as we know Q(s)-P(s) = 5
17(t+2) - 5(t+2)^2 -(7t-5t^2) = 5
Solve this and the value of t would be 0.9s.

Just use v = u + at now. Use t+2 for P and t for Q. You will get the answer

From where did the fives from s = 17(t+2) - 5(t+2)^2 and s = 7t - 5t^2 come from?
 
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8GN0L.png


Can't figure this gem out. Any help, please?
 
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Last edited:
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View attachment 41973
plz explain iii and iv
i could explain it to you. it would take the same time as it took me to search for this video. but actually this video will build the concept and explain in a better way. so i am giving you this link not because i am lazy but because it explains in a better way.
http://www.examsolutions.net/maths-...ertical-strings/hitting-ground/tutorial-1.php
if youtube is blocked then use any VPN or use this website http://www.fattol.com
 
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