Question

The relation between time $t$ and displacement $x$ is $t=\alpha x^2+\beta x$, where $\alpha$ and $\beta$ are constant, find the relation between velocity and acceleration.

Answer 1

$t=\alpha x^2+\beta x$ .......................1

Differentiaitng above equation wrt $x$ we get $\left(dt/dx\right)=2\alpha x+\beta$ ..............................2

but $velocity\left(v\right)=dx/dt$ so $v=\frac{1}{2\alpha x+\beta }$

differentiaitng equation 2 wrt x we get$\left(d^{2}t/d{x}^2\right)=2\alpha$

As $acceleration\left(a\right)=d^{2}x/d{t}^2$ so we have $a=1/2\alpha$...................3

so we have $2\alpha =1/a$

put it in expression of velocity. Thus we get $v=\frac{1}{\frac{x}{a}+\beta }$.

It is the required relation between velocity and acceleration/