Question

Two particles \(A\) and \(B\) move from rest along a straight line with constant accelerations \(f\) and \(h\) respectively. If \(A\) takes \(m\) seconds more than \(B\) and describes \(n\) units more than that of \(B\) acquiring the same speed, then
(A) \((f+h) m^{2}=f h n\)
(B) \((f-f h) m^{2}=f h n\)
(C) \((h-f) n=\frac{1}{2} f h m^{2}\)
(D) \(\frac{1}{2}(f+h) n=f h m^{2}\)

Answer 1

Ans: (C)
Hint \(: S+n=\frac{1}{2} f(t+m)^{2}\) and \(S=\frac{1}{2} h t^{2}, V=h t\)
$$
\begin{aligned}
&\therefore \frac{1}{2} h t^{2}+n=\frac{1}{2} f(t+m)^{2}-(1) \\
&\text { Also } V=0+h t=0+f(t+m) \quad \Rightarrow t+m=\frac{h t}{f}
\end{aligned}
$$
From equation (1),
$$
\frac{1}{2} h t^{2}+n=\frac{1}{2} f\left(\frac{h t}{f}\right)^{2} \Rightarrow t^{2}=\frac{2 n f}{h(h-f)}
$$
Also,
$$
\begin{aligned}
&h t=f(t+m) \quad \Rightarrow t^{2}=\frac{m^{2} f^{2}}{(h-f)^{2}} \\
&\therefore \frac{2 n f}{h(h-f)}=\frac{m^{2} f^{2}}{(h-f)^{2}} \quad \Rightarrow 2 n=\frac{m^{2} f h}{h-f} \quad \Rightarrow n(h-f)=\frac{1}{2} f h m^{2}
\end{aligned}
$$

Answer 2

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