Question

A point particle starts moving in straight line with const. acc. "a" with time "t1".After the beginning of the motion,acc changes its sign,remaining same in magnitude.Time t in the beginning of the motion i nwhich particle returns to its initial position will be ______?

Answer 1

TIME WILL BE 4T BECAUSE ACCN AND RETARDATION IS CARRIED OUT AT SAME RATE LET US SAY THAT IT COVERS "S" DISTANCE WHILE XLR8ING THEN AFTER TIME T IT WOULD AGAIN TRAVEL A DISTANCE "S"WHILE RETARDING BECAUSE VELOCITY IS IN OPP DIRECTION TO RETARDATION.WHEN THE RETARDATION WOULD OVERPOWER THE INITIAL VELOCITY CHANGING THE DIRECTION OF ITS VELOCITY THEN FINALLY IT WILL HAVE TO COVER A DISTANCE "2S''HENCE NEEDING A TIME OF "2T" MORE TO RETURN ITS ORIGINAL POSITION

Answer 2

Please don't answer in all capital letters.

To answer the question, I must first clarify it. I assume that the particle is initially at rest, and at time t= 0 begins accelerating. At time t1, the sign of the acceleration switches. The question asks the time at which the particle returns to its starting point.


If that's the case, then the distance covered in the time t1 is
\[ \Delta x = \frac{1}{2}at_1^2\]
and the velocity at time t1 will be
\[ v = at_1 \]
At this point, the acceleration changes sign. The time it takes for the particle to stop is simply t1 again. The distance covered between t = t1 and t= 2t1 is
\[\Delta x = v_0t + \frac{1}{2}at^2 = at_1^2 - \frac{1}{2}at_1^2 = \frac{1}{2}at_1 ^2\]
which could have also been obtained by considering symmetry. Therefore, the particle is again at rest, and is displaced from its starting point a total distance of
\[ \Delta x = at_1^2 \]
which we obtain by adding the two displacements found above.

Now we ask how long it will take to return to the starting point. We can simply use the equation
\[\Delta x = \frac{1}{2}at^2 = at_1^2\]
solving for t, we find that
\[t = t_1\sqrt{2}\]
Therefore, adding all of those times together, it will return to the origin at
\[t = t_1 + t_1 + t_1\sqrt{2} = (2+ \sqrt{2}) t_1\]

Answer 3

Thanks!