Question

a particle Q travels a straight line with a displacement of 43m. velocity of 41m/s and an accelaration of 36(m/sec^2) and time in 2sec.find the equation of the line.....

Answer 1

Can you be more clear? Velocity at the start is 41m/s or at the end? Is the acceleration constant?

Answer 2

but frm s =ut +1/2at^2 displacement does comes out to b 41 m..

Answer 3

without using constatnt acceleration formula to form the equation..
i think derivative or integration is to be use to form the equation...

Answer 4

I have no idea what is going on, so I will give you the general way to solve it.
We start with acceleration, which is the rate of change of velocity, which is itself the rate of change of displacement, hence:
\[a=\frac{d}{dt}(\frac{dx}{dt})=\frac{d^{t}x}{dt^{t}}\]\[Integrating\:both\:sides: v-v_{0}=\int_{0}^{t}a dt=at\]\[v=at+v_{0} \:\:(1)\]Integrating both sides again:
\[\int_{0}^{t}\frac{dx}{dt}dt=\int_{0}^{t} (at-v_{0})dt\]
This gives:
\[x-x_{0}=\frac{1}{2}at^{2}+v_{0}t\]
Thus we get:
\[x=x_{0}+v_{0}t+\frac{1}{2}at^{2}\:\:(2)\]

Now using equation (1):
\[41=36(2)+v_{0}\]\[v_{0}=-31ms^{-2}\]
Then using equation (2):
\[43=x_{0}+\frac{36}{2}2^{2}-31(2)\]
\[x_{0}=33m\]
So the final expression is:
\[x(t)=33-31t+18t^{2}\]

There are lots of other assumptions I have not included, which you have not included either in the question.