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 total time in which the point returns to the initial position is

• A. $\sqrt{2t}$
• B. $(2+\sqrt{2})t$
• C. $\frac{1}{\sqrt{2}}$
• D. Cannot be predicted unless acceleration is provided

Answer 1

The correct option is B

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

In this problem point starts moving with a uniform acceleration a and after time t (Position B) the direction of acceleration get reversed i.e, the retradation of same value works on the point.Due to this velocity of the point goes on decreasing and at position C its velocity becomes zero.Now the direction of motion of point reversed and it moves from C to A under the effect of acceleration a.

We have to calculate the total time in this motion.Starting velocity at position A is equal to zero. Velocity at position B$\Rightarrow v=at$ [As u = 0]

Distance between A and B, $SAB=\frac{1}{2}a{t}^2$

As same amount of retardation works on a point and it comes to rest therefore $S_{BC}=S_{AB}=\frac{1}{2}a{t}^2$

$\therefore$ Total time taken for motion between A and C = 2t

Now for the return journey from C to A$(S_{AC}=a{t}^2)$

$S_{AC}=ut-\frac{1}{2}a{t}^2\Rightarrow -a{t}^2=0-\frac{1}{2}a{t}_1{}^2\Rightarrow t_1=\sqrt{t}$

Hence total time in which point returns to initial point T = $2t+\sqrt{2}t=(2+\sqrt{2})t$