Question

A point mass starts moving in a straight line with constant acceleration 'a'. At a time t after start the acceleration changes sign remaining same in magnitude find time 'T' from beginning of motion in which the point returns to original position:

• A. $t(\sqrt{2}-1)$
• B. $t(\sqrt{2}+2)$
• C. $t(\sqrt{2}+1)$
• D. $t(2-\sqrt{2})$

Answer 1

The correct option is B $t(\sqrt{2}+2)$

The correct option is B.

Given,

A point mass starts with accelertion $a$

Equations of motion are:

$s=ut+\frac{1}{2}a\times t^2$,

$u=0$ and at time $t_1,$ when the acceleration changes, distance travelled:

$s=\frac{1}{2}at{1}^2$

$vatt=t_1=u+at=a{t}_1$

Now the acceleration is changed to -a. Then the particle continues in the same direction until the velocity becomes zero. Then the particle changes the direction and starts accelerating and passes over the point of start.

$u=a{t}_{1}v=0$ acceleration $=-a$

$v=u+at$

$\Rightarrow 0=a{t}_1-at$

$\Rightarrow t=t_1$ it takes $t_1$ more time to stop and reverse direction.

The distance traveled/displacement in this time:

$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow s=a{t}_1\times t_1-\frac{1}{2}a{t}_1^2=\frac{1}{2}a{t}_1^2$

The total displacement from the initial point: $\frac{1}{2}a{t}_1^2+\frac{1}{2}a{t}_1^2=a{t}_1^2$

Now,

acceleration $=-a$ $u=0$ $s=-a{t}_1^2$ in the negative direction

$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow -a{t}_1^2=0-\frac{1}{2}a{t}^2$

$\Rightarrow t=\sqrt{2}{t}_1$

The total time T from initial point forward till back to initial point :

$T=2t_1+\sqrt{2}{t}_1=(2+\sqrt{2})t_1$