Question

Displacement $X$ of a particle at time $t$ moving along a straight-line path is given by $X^2=a{t}^2+2bt+c$, where $a,b$ and $c$ are constants. If acceleration of the particle varies as $\frac{a-v^n}{X^m}$ , where $v$ is velocity of particle at time 𝑡, then find $(n+m)$

Answer 1

Given:
$X^2=a{t}^2+2bt+c...\left(i\right)$
By differentiating the equation (i), we get
$2X\frac{dX}{dt}=2at+2b2xv=2at+2b...\left(ii\right)$
differentiating equation $\left(ii\right)$ with request to time, we get
As we know,
$v=\frac{dx}{dt}a=\frac{dv}{dt}$
Therefore,
$v^2+xA=aA=\frac{a-v^2}{x}...\left(iii\right)$
Where A is the acceleration at any instant $t$.
Acceleration of the particle varies as $\frac{a-v^n}{X^m}...\left(iv\right)$
Comparing equation $\left(iii\right)$ and $\left(iv\right)$ we get,
$n=2,m=1n+m=2+1=3$