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Mathematics: Post your doubts here!

RoOkaYya G

RoOkaYya G

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Abdel Moniem said:
Cyclist P reached B at speed 5 m/s and travelled a distance from B to C of 1400 m

Cyclist P moves with constant speed of 5 m/s t=1400/5=280 seconds to travel from B to C. at t=400 second the speed of cyclist Q is 4/3m/s this represents the intial velocity. plug that into the equation below
use s=ut+1/2at^2
1400=4/3*280+0.5(280)^2a
Click to expand...
he speed of cyclist Q is 4/3m/s ====> how u got this? thts wht i wanna knw
 
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Abdel Moniem

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Thought blocker said:
How is speed of cyclist Q 4/3?
Click to expand...
Cyclist Q starts from t=400 s and so on to move with constant acceleration. now, find the speed of Q at 400 seconds from the equation given v=0.04t-0.0001t^2+k. speed of Cyclist Q represent the initial velocity( from point B to C) which increases as time passes.
v(t)=0.04t-0.0001t^2+k
v(400)=0.04(400)-0.0001(400)^2+k=4/3
Note: we dont use this equation v(t)=0.04t-0.0001t^2+k anymore as it belongs to only part A to B.
 
Thought blocker

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Abdel Moniem said:
Cyclist Q starts from t=400 s and so on to move with constant acceleration. now, find the speed of Q at 400 seconds from the equation given v=0.04t-0.0001t^2+k. speed of Cyclist Q represent the initial velocity( from point B to C) which increases as time passes.
v(t)=0.04t-0.0001t^2+k
v(400)=0.04(400)-0.0001(400)^2+k=4/3
Note: we dont use this equation v(t)=0.04t-0.0001t^2+k anymore as it belongs to only part A to B.
Click to expand...
I solved the question, and I got it. :D
Thanks though.
I guess, Rookz Adulto, never solved the other parts. :D:D:D
 
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RoOkaYya G said:
yh but how does initial speed u is 4/3 ???
4/3 of part ii is value of k whihc is a constant .-.
Click to expand...
that was just a coincidence that at t=400 seconds speed was equal to k. in the equation equation for v, t lasts only for 400 seconds as it says in the question that both cyclists reach point B at 400 seconds. Equation is for part A to B only. now we can find the speed of cyclist Q at the 400th second by plugging in t=400 into the equation given. speed at B is the speed from which the cylcist starts to move(initial speed) and gradually increases because of constant acceleration
 
RoOkaYya G

RoOkaYya G

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Abdel Moniem said:
that was just a coincidence that at t=400 seconds speed was equal to k. in the equation equation for v, t lasts only for 400 seconds as it says in the question that both cyclists reach point B at 400 seconds. Equation is for part A to B now we can find the speed of cyclist Q at the 400 second by plugging in t=400 into the equation given. speed at B is the speed from which the cylcist starts to move(initial speed) and gradually increases because of constant acceleration
Click to expand...
v(t)=0.04t-0.0001t^2+k
v(400)=0.04(400)-0.0001(400)^2+k=4/3

u wrote this above. how u got the v multiplying with t? (in green)
 
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