Question

block starts moving on a horizontal plane under the influence of constant accelerations a. After t sec, the acceleration
remains the same in magnitude but reverses in direction. At what time does it come back to the wil point​

Answer 1

Answer:

its to fast I will also clarify it

Answer 2

Given:

Blocks start moving on a horizontal plane under the influence of constant acceleration a. After t sec, the acceleration remains the same in magnitude but reverse in direction.

To Find:

We are asked to find the time at which it come back to the will point.

Solution:

Using first equation of motion,

At time $t_{0}$,

$v=u+at_{0} =at_{0}$

Now acceleration is in opposite direction, the total distance covered by particle is given by, $2*(\frac{1}{2}*a*t_{0} ^{2})=at_{0} ^{2}$.

After covering this distance, the particle moves in reverse direction with zero initial velocity.

$v^{2}=0+2as$      [ $s=at_{0} ^{2}$ ]

$v^{2}=2*a*(at_{0} ^{2})$

$v=\sqrt{2at_{0} }$

total time is given by, $t_{0}+t_{0}+\sqrt{2t_{0} } =t_{0}(2+\sqrt{2})$

Hence time at which block come back to will point is $t_{0}(2+\sqrt{2})$.

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