Question

Two fixed points $A$ and $B$ are $20$ metres apart. At time $t=0$, the distance between a third point $C$ and $A$ is $20$ meters and the distance between $C$ and $B$ is $10$ metres. The component of velocity of point $C$ along both $CA$ and $CB$ at any instant is $5m/s$. Then the distance between $A$ and $C$ at the instant all the three points are collinear will be

• A. $5m$
• B. $15m$
• C. $10m$
• D. $25m$

Answer 1

Answer: (B)

The length of side $CA$ at any time $t$ is $=20-5t$.
The length of side $CB$ at any time $t$ is $=10-5t$.
At the instant $A,B$ and $C$ are collinear
$(20-5t)+(10-5t)=20$.
Solving we get $t=1$.
Therefore, length of $CA$ at $t=1$ is $20-5=15m$