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I am given $v=6-4x$ where $v$ is velocity and $x$ is position of the body. I am now asked to find displacement between $t=1.4$, distance covered from $t=1.4$, average velocity from $t=2.5$, average acceleration, etc. I am not able to find an approach to these kind of problems where $v$ is not given in terms of $t$, or say, it is given like $v=kt^2$ or like $v=kx$. How should I start these kind of problems and develop a thinking for these? Please help me.

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First thing to do is to write $v = dx/dt$. That's by definition. Your formula is now $dx/dt = 6-4x$. This is a separable differential equation. Solving this yields an expression for $x$ in terms of $t$. The rest of the problem can now be solved.

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  • $\begingroup$ Thanks @Allure but i am still confused for avg distance and avg speed, i got x=3/2-1/(2e^4t). It was given at t=0, x=1 $\endgroup$ – FastAndTheCurious Jun 16 at 3:02
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    $\begingroup$ @AyushBhardwaj once you have $x$ as a function of $t$, you can calculate the average velocity between (say) $t=0$ and $t=1$ by calculating where the particle is at those times, and dividing by the time taken to get there (1s in this case). $\endgroup$ – Allure Jun 16 at 3:09
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    $\begingroup$ Remember that the average value of some function $f(x)$ on $[a,b]$ is $$\bar f_{[a,b]}=\dfrac1{b-a}\int_a^b f(x) \ \mathrm dx$$ $\endgroup$ – user256872 Jun 16 at 3:11
  • $\begingroup$ @AyushBhardwaj The average speed is the distance traversed over time; the average of $x=(6-v)/4$ is $(6-\bar{v})/4$. $\endgroup$ – J.G. Jun 16 at 6:07

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