Question

A particle starts with initial speed $u$ and retardation $a$ comes to rest in time $t$. The time taken to cover first half of the total path travelled is:

Answer 1

We can solve this question by using the three equations of motion. These equations are used to describe the nature of motion of any given system. Assuming that the particle travels in a straight line and the acceleration is constant, we can solve the sum.

Formula used: $v=u+at,$
$s=ut+\dfrac{1}{2}at^{2},$
$v^{2}-u^{2}=2as$
Where, $u,\;v$ is the initial and the final velocity of the particle, which covers a distance $s$ in time $t$ with an acceleration $a$.

Complete step by step answer:
Given that the initial speed of the particle is $u$ , the retardation of the particle is $a$ and the time taken for the particle to come to rest is $t$. Let us assume that the particle is travelling in the straight line and the acceleration is constant. Let $v$ be the final velocity of the particle and $s$ is the total distance covered by the particle. Then let the particle take time $T$ taken to cover a distance $S=\dfrac{s}{2}$.
From the equation of motion, we have
$v=u+at,$
Substituting the values, in the respective places we get,
$0=u-at$
Since acceleration is retarding we write as $-a$. On rearranging, we get, the total time $t$ as
$\implies t=\dfrac{u}{a}$
Also, we know that
$v^{2}-u^{2}=2as$
Substituting for the values, we get, the total distance covered $s$ as
$s=\dfrac{(0-u^{2})}{-2a}$
$\implies s=\dfrac{u^{2}}{2a}$
Then, from
$s=ut+\dfrac{1}{2}at^{2},$
We can find the time taken $T$ to cover a distance $S$ ,substituting for the values, we get,
$\implies S=uT-\dfrac{1}{2}aT^{2}$
Then since $S=\dfrac{s}{2}$, we get,
$\implies \dfrac{s}{2}=uT-\dfrac{1}{2}aT^{2}$
Substituting for $s$, we get$\implies \dfrac{u^{2}}{4a}=uT-\dfrac{1}{2}aT^{2}$
$\implies u^{2}=4uaT-2(aT)^{2}$
$\implies 2(aT)^{2}-4uaT+u^{2}=0$
This is similar to quadratic equation in terms of $T$
Then, $T=\dfrac{4ua\pm\sqrt{(4ua)^{2}-4.2a^{2}.u^{2}}}{2.2a^{2}}$
$\implies T=\dfrac{4ua\pm\sqrt{16(ua)^{2}-8(ua)^{2}}}{4a^{2}}$
$\implies T=\dfrac{4ua\pm\sqrt{8(ua)^{2}}}{4a^{2}}$
$\implies T=\dfrac{4au\pm2au\sqrt2}{4a^{2}}$
$\implies T=\dfrac{2ua(2+\sqrt 2)}{ 4a^{2}}\;or\;\dfrac{2ua(2-\sqrt 2)}{ 4a^{2}}$
$\implies T=\dfrac{u(2+\sqrt2)}{2a}\;or\;\dfrac{u(2-\sqrt2)}{2a}$
But we have $\implies t=\dfrac{u}{a}$
On simplification, we get, $T=t(2+\sqrt 2)\;or\;t(2-\sqrt2)$
Since time taken to cover half distance is less than the time taken for the particle to cover full distance, we can say that $T=t(2-\sqrt 2)$
Therefore, the time taken to cover first half of the total path travelled is $T=t(2-\sqrt 2)$

Note: The sum involves only the rearrangement of the three equations of motion. Also the solution of the quadratic equation $ax^{2}+bx+c=0$ in terms of $x$ is given as $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. We are using this in the above sum.