Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

A man can swim in still water with a speed of $3\dfrac{m}{s}$, x and y axis are drawn along and normal to the bank of river flowing to right with a speed of $1\dfrac{m}{s}$. The man starts swimming from origin O at $t = 0s$. Assume the size of man to be negligible. Find the equation of locus of all the possible points where man can reach at $t = 1s$.

Answer
VerifiedVerified
466.5k+ views
Hint: A man can swim in still water with a speed of three metres per second, so we can say that the man is swimming a distance of three metres in one second. Similarly, one metre per second can be said as one metre distance travelled in one second. With the help of this concept, we will find the coordinates of the given point and hence, we will get the equation of the locus.

Complete step by step answer:
Considering that the river is still, then the man would have travelled a distance of $3m$ from origin $O$ after a time period of $1\sec $. The locus of any point at which the man can reach at $t = 0s$ is a semicircle which is of a radius $3m$ and the centre is present at $O$.
We know that the river flows towards the right at a speed of $1\dfrac{m}{s}$.
Therefore, there should be an additional shift in position by $1\dfrac{m}{s}$, that is $1\sec = 1m$ towards the right. So, the locus of every point giving possible position after a time period of one second will be a semicircle shifted to right by $1m$.
Thus, the locus of every point where the man can be at $t = 1s$ is also a semicircle which is of a radius $t = 1s$ and centres at ${O'}(1m,0m)$.
seo images

Hence, the equation of locus of all the points comes out to be,
${(x - 1)^2} + {(y - 0)^2} = {3^2}$
On simplifying the above equation, we get,
${(x - 1)^2} + {y^2} = 9$
The equation of locus of all the possible points where man can reach at $t = 1s$ is ${(x - 1)^2} + {y^2} = 9$ which is a circle.

Note:
 The above equation ${(x - 1)^2} + {y^2} = 9$ is the equation of a circle. By equating it with the standard equation of the circle, we will get the radius of this circle as $3m$ and the centre is present at the point $(1,0)$.