Question

A point starts moving in a straight line with a certain acceleration. At a time t after beginning of motion the acceleration suddenly becomes retardation of the same value. The time in which the point returns to the initial point is

• A. $\sqrt{2t}$
• B. $(2 + \sqrt{2})t$
• C. $\frac{t}{\sqrt{2}}$
• D. Cannot be predicted unless acceleration is given

Answer 1

Answer: (B)

$(2 + \sqrt{2})t$

Answer 2

Let the initial position of the point be $x_0 = 0$ and the initial velocity be $v_0 = 0$. The motion is divided into two parts.

Part 1: Motion with acceleration $a$ for time $t$.

• Initial velocity $u_1 = 0$.
• Acceleration $a_1 = a$.
• Duration $\Delta t_1 = t$.

The velocity at the end of this part is $v_1 = u_1 + a_1 \Delta t_1 = 0 + at = at$.
The displacement during this part is $s_1 = u_1 \Delta t_1 + \frac{1}{2} a_1 (\Delta t_1)^2 = 0 \cdot t + \frac{1}{2} a t^2 = \frac{1}{2} a t^2$.
At time $t$, the position of the point is $x_1 = x_0 + s_1 = \frac{1}{2} a t^2$.

Part 2: Motion with retardation of the same value, starting from time $t$.

• The initial velocity for this part is $u_2 = v_1 = at$.
• The acceleration for this part is $a_2 = -a$.
• Let the duration of this part be $\Delta t_2$.

The point returns to the initial point, so the final position is $x_f = x_0 = 0$.
The displacement during Part 2 is $s_2 = x_f - x_1 = 0 - \frac{1}{2} a t^2 = -\frac{1}{2} a t^2$.

Using the kinematic equation for displacement in Part 2:
$s_2 = u_2 \Delta t_2 + \frac{1}{2} a_2 (\Delta t_2)^2$
$-\frac{1}{2} a t^2 = (at) \Delta t_2 + \frac{1}{2} (-a) (\Delta t_2)^2$
$-\frac{1}{2} a t^2 = at \Delta t_2 - \frac{1}{2} a (\Delta t_2)^2$

Assuming $a \neq 0$, we can divide the entire equation by $a$:
$-\frac{1}{2} t^2 = t \Delta t_2 - \frac{1}{2} (\Delta t_2)^2$

Rearrange the terms to form a quadratic equation in $\Delta t_2$:
$\frac{1}{2} (\Delta t_2)^2 - t \Delta t_2 - \frac{1}{2} t^2 = 0$
Multiply by 2:
$(\Delta t_2)^2 - 2t \Delta t_2 - t^2 = 0$

This is a quadratic equation for $\Delta t_2$. Using the quadratic formula $\Delta t_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$, $b=-2t$, $c=-t^2$:
$\Delta t_2 = \frac{-(-2t) \pm \sqrt{(-2t)^2 - 4(1)(-t^2)}}{2(1)}$
$\Delta t_2 = \frac{2t \pm \sqrt{4t^2 + 4t^2}}{2}$
$\Delta t_2 = \frac{2t \pm \sqrt{8t^2}}{2}$
$\Delta t_2 = \frac{2t \pm 2t\sqrt{2}}{2}$
$\Delta t_2 = t \pm t\sqrt{2}$
$\Delta t_2 = t(1 \pm \sqrt{2})$

Since $\Delta t_2$ represents a duration of time, it must be positive.
So, the valid duration for Part 2 is $\Delta t_2 = t(1 + \sqrt{2})$.

The total time taken for the point to return to the initial point is the sum of the durations of Part 1 and Part 2:
Total time $T = \Delta t_1 + \Delta t_2 = t + t(1 + \sqrt{2})$
$T = t + t + t\sqrt{2}$
$T = 2t + t\sqrt{2}$
$T = (2 + \sqrt{2})t$.