A particle moving with uniform retardation along a straight line covers distances $a$ and $b$ in successive intervals $p$ and $q$ seconds. The acceleration of the particle is

Question

Aicharya · Accepted Answer

Let retardation be $f$ and initial velocity be $u$. For $A B$, $\begin{aligned} & a=u p-\frac{1}{2} f p^2 \quad \ldots (i) \ & \text {For } A C, a+b=u(p+q)-\frac{1}{2} f(p+q)^2 \quad \ldots (ii) \ & \text {(i) and (ii) } \Rightarrow a+b \ & =\frac{\left(a+\frac{1}{2} f p^2\right)}{p}(p+q)-\frac{1}{2} f(p+q)^2 \ & \frac{a+b}{p+q}=\frac{a}{p}+\frac{1}{2} f p-\frac{1}{2} f p-\frac{1}{2} f q, \ & \frac{1}{2} f q=\frac{a}{p}-\frac{a+b}{p+q} \quad=\frac{a p+a q-a p-b p}{p(p+q)} \ & \frac{1}{2} f q=\frac{a q-b p}{(p+q)}, \quad f=\frac{2(a q-b p)}{\mathrm{p} q(p+q)} \end{aligned}$

A particle moving with uniform retardation along a straight line covers distances $a$ and $b$ in successive…

Solution