During the shooting of a superhit film 'MARD' Amitabh Bachchan was waiting for his beloved Amrita Singh with his dog. When he saw her approaching, the dog was excited and dashed to her then went back to his master and so on, never stopping. How far would you estimate the dog ran, if its speed is 30 km/h and each of them walked at 4 km/h, starting 400 m apart? If your answer is $s = 5x \times {10^2}\,m$, find the value of $x$.
Answer
280.5k+ views
Hint: In this solution, we will use the concept of relative velocity. Since Mr. Amitabh and Ms. Amrita are moving towards each other, their relative velocity will be the sum of their individual velocity.
Complete step by step answer:
We’ve been told that Mr. Amitabh and Ms. Amrita are walking towards each other at a speed of 30km/h and the dog is running between them.
We will first calculate the time taken by Mr. Amitabh and Ms. Amrita to meet each other. For this, we will use the concept of relative velocity. Since they are walking towards each other, their relative velocity given to us is the sum of their individual velocity. Since they are both walking at a speed of 4km/h, their net relative velocity will be
${v_{AA}} = 4 - ( - 4)$
$ \Rightarrow {v_{AA}} = 8\,{\text{km/h}}$
Hence the time they will take to cover the distance between them will be determined from the relation of speed, distance and time as
$t = \dfrac{d}{{{v_{AA}}}}$
Since the distance between them is $d = 400\,m = 0.4\,km$, we can calculate
$t = \dfrac{{0.4}}{8}$
$ \Rightarrow t = 0.05\,hr$
Now, in this time, the dog will be running back and forth between Mr. Amitabh and Ms. Amrita. So, the distance it will cover travelling for $t = 0.05\,hr$ at a speed of ${v_D} = 30\,km/h$ will be
$d = {v_D}t$
$ \Rightarrow d = 30 \times 0.05$
Which gives us
$d = 1.5\,km = 1500\,m$
Since we’ve been given the distance in the question as $s = 5x \times {10^2}\,m$, the value of $x$ can be calculated using
$5x \times {10^2} = 1500$
Hence,
$x = 3$
Note: The trick to solving such questions to find out the time for which the dog moves for using the data given in question. We might make the mistake of calculating the distance travelled by the dog when it moves back and forth but it will be too tedious and instead we can use the concept of relative velocity to determine the time it moves for and then calculate the distance directly as we did in the solution above.
Complete step by step answer:
We’ve been told that Mr. Amitabh and Ms. Amrita are walking towards each other at a speed of 30km/h and the dog is running between them.
We will first calculate the time taken by Mr. Amitabh and Ms. Amrita to meet each other. For this, we will use the concept of relative velocity. Since they are walking towards each other, their relative velocity given to us is the sum of their individual velocity. Since they are both walking at a speed of 4km/h, their net relative velocity will be
${v_{AA}} = 4 - ( - 4)$
$ \Rightarrow {v_{AA}} = 8\,{\text{km/h}}$
Hence the time they will take to cover the distance between them will be determined from the relation of speed, distance and time as
$t = \dfrac{d}{{{v_{AA}}}}$
Since the distance between them is $d = 400\,m = 0.4\,km$, we can calculate
$t = \dfrac{{0.4}}{8}$
$ \Rightarrow t = 0.05\,hr$
Now, in this time, the dog will be running back and forth between Mr. Amitabh and Ms. Amrita. So, the distance it will cover travelling for $t = 0.05\,hr$ at a speed of ${v_D} = 30\,km/h$ will be
$d = {v_D}t$
$ \Rightarrow d = 30 \times 0.05$
Which gives us
$d = 1.5\,km = 1500\,m$
Since we’ve been given the distance in the question as $s = 5x \times {10^2}\,m$, the value of $x$ can be calculated using
$5x \times {10^2} = 1500$
Hence,
$x = 3$
Note: The trick to solving such questions to find out the time for which the dog moves for using the data given in question. We might make the mistake of calculating the distance travelled by the dog when it moves back and forth but it will be too tedious and instead we can use the concept of relative velocity to determine the time it moves for and then calculate the distance directly as we did in the solution above.
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