Question

A tiger chases a deer $30m$ ahead of it and gains $3m$ in $5s$ after the chase begins. The distance gained by the tiger in $10s$ is
A.$6m$
B.$12m$
C.$18m$
D.$20m$

Answer 1

Hint: To answer this question we have to assume that initially, the tiger was at rest. Now, we will use the second equation of motion to find the acceleration of the given distance and time. After getting the acceleration, we will again assume that the acceleration is uniform i.e. it remains constant throughout the run and again using the second equation of motion, we will get the distance in the desired time.
Formula Used:
$s = ut + \dfrac{1}{2}a{t^2}$
Here,
$s$ is the displacement.
$u$ is the initial velocity.
$t$ is the time taken.
$a$ is the acceleration.

Complete answer:
Let us assume that the tiger is at rest initially. So the initial velocity of tiger,
$u = 0m/s$
Now, in the question, it is given that the deer is $30m$ ahead of the tiger. So, the initial distance between the tiger and deer is:
$d = 30m$
But, we only know the distance covered in $5s$. So, we will use the second equation of motion to find acceleration using the distance $3m$ covered in time $5s$.
The Second Equation of Motion is given below:
$s = ut + \dfrac{1}{2}a{t^2}......(1)$
Here,
$s$ is the displacement.
$u$ is the initial velocity.
$t$ is the time taken.
$a$ is the acceleration.
Now, as the tiger is running and chasing the tiger, this implies that it is accelerating. To find acceleration we have to put the given values in the second equation of motion, we get,
$u = 0m/s$
$s = 3m$
$t = 5s$
$s = ut + \dfrac{1}{2}a{t^2}$
$ \Rightarrow 3 = 0 \times 5 + \dfrac{1}{2}a \times {5^2}$
$ \Rightarrow 6 = a \times 25$
$ \Rightarrow a = \dfrac{6}{{25}}m/{s^2}$
Now, we have to find the distance covered by the tiger in $10s$. Here, we assume that the tiger is accelerating in a uniform manner i.e. the acceleration is constant throughout the chase. We have to put the value of acceleration we get above into equation (1), we get,
$t = 10s$
$s = ut + \dfrac{1}{2}a{t^2}$
$ \Rightarrow s = 0 \times 10 + \dfrac{1}{2} \times \dfrac{6}{{25}} \times {10^2}$
$ \Rightarrow s = \dfrac{1}{2} \times \dfrac{6}{{25}} \times 100$
$ \Rightarrow s = 12m$

So, option B is the correct answer.

Note:
Whether an object is in motion or at rest is determined by a reference point. If the object is changing its position with respect to a reference point then it is said to be in motion and if the position remains the same with respect to the reference point then it is at rest. The three equations of motion can only be applied if the acceleration of the body moving is constant with time and the body moves in a straight path.