Question

Point ${}^{\prime }{A}^{\prime }$ moves uniformly with speed $v_1(=20m/sec)$ so that vector ${\overrightarrow{v}}_1$ is continuously 'aimed' at point ${}^{\prime }{B}^{\prime }$ which in turn moves rectilinearly and uniformly with velocity $v_2(=10m/sec)$ along the path $P\to Q$ as shown in the figure. If the time (in sec)when the points A and B converge is $\frac{3k}{7}$. Then, find the value of k?

Answer 1

Relative velocity of $A$ with respect to $B$ along the line of approval
$\Rightarrow V_1-V_2\cos \theta =-\frac{dl}{dt}$
${\int }_0^t(V_1-V_2\cos \theta )dt={\int }_l^0-dl$
$V_{1}t-V_2{\int }_0^t\cos \theta dt=l$ (equation $1$)
The distance covered along $B{B}^{\prime }$ when $A$ converges with $B$ is equal
$\therefore {\int }_0^t{V}_1\cos \theta dt=V_{2}t$
$\Rightarrow {\int }_0^t\cos \theta dt=\frac{V_{2}t}{V_1}$ (equation $2$)
Substituting $2$ in $1$
$V_{1}t-V_2\left(\frac{V_{2}t}{V_1}\right)=l$
$t(V_1^2-V_2^2)=l$
$t=\frac{l}{V_1^2-V_2^2}=\frac{300/7}{400-100}=\frac{1}{7}=\frac{3k}{7}$
$K=\frac{1}{3}$