Question

On a large slippery ground, a boy left his dog sitting and walked away with constant velocity ${v_b} = 2m/s$. When he reaches ${x_0} = 199m$ away from the dog, the dog decides to catch him and meet at the same speed as that of the boy. Because the ground is slippery, the dog cannot develop acceleration more than $a = 2m/{s^2}$ in any direction. In what minimum time will the dog meet the boy?

Answer 1

We have the values of velocity of the boy and the dog, distance to be covered by the dog and the maximum value of acceleration. By using the equations of motion, we can calculate the minimum time in which the dog will meet the boy.

Formula used:
The equations of motion are given as
$v = u + at \\\ S = ut + \dfrac{1}{2}a{t^2} \\\ {v^2} - {u^2} = 2aS \\\$

Complete step by step answer:
The velocity with which the boy and the dog move is same and its value is given as
${v_b} = 2m/s$
Also the initial velocity of the dog is zero since the dog starts moving from rest position.
$u = 0$
The distance travelled by the dog to reach the boy is given as
${x_0} = 199m$
But the boy is still walking away from the dog so the dog will have to travel an additional distance which is given as the product of velocity and time taken by the dog to reach the boy. Let this time be t, then total distance covered by dog is given as
$S = {x_0} + {v_b}t = 199 + 2t$
The maximum acceleration that the dog can develop on the slippery ground is given as
$a = 2m/{s^2}$
Now, let us insert the above values of acceleration, velocity and distance in the second equation of motion. Doing so, we get
$S = ut + \dfrac{1}{2}a{t^2} \\\ 199 + 2t = 0 + \dfrac{1}{2} \times 2{t^2} \\\ {t^2} - 2t - 199 = 0 \\\$
Now solving for t, we get
$t = \dfrac{{2 \pm \sqrt {4 + 4 \times 199} }}{2} \\\ = \dfrac{{2 \pm 28.28}}{2} \\\ = \dfrac{{2 + 28.28}}{2},\dfrac{{2 - 28.28}}{2} \\\ = 15.14, - 13.14 \\\$
Since the value of time cannot be negative, the correct value of time taken by a dog to reach the boy is 15.14 seconds. This is the required answer.

Note:
It should be noted that the boy is still moving when the dog decides to reach him and the boy has travelled an additional distance while the dog tries to reach him. Without taking this information into consideration, the student will end up getting the wrong answer.