Question

A point particle starts moving in a straight line with constant acceleration. After tome t0 the

acceleration of the particle is reversed, magnitude remaining the same. Determine the time

from the beginning of the motion in which the particle returns to the initial position.


Ans:
$4t0$
​

Answer 1

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:

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

$vaat = t1 = u + at = at1$

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.

$u = at1 \: v = 0 \: acceleration \: = - a$

$v = u + at$

$\: \: \: 0 = at1 - at$

$t = t1$

it takes t

1

more time to stop and reverse direction.

The distance traveled/displacement in this time:

s=ut+

2

1

at

2

⇒s=at

1

×t

1

−

2

1

at

1

2

=

2

1

at

1

2

The total displacement from the initial point:

2

1

at

1

2

+

2

1

at

1

2

=at

1

2

Now,

acceleration =−a u=0 s=−at

1

2

in the negative direction

s=ut+

2

1

at

2

⇒−at

1

2

=0−

2

1

at

2

⇒t=

2

t

1

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

T=2t

1

+

2

t

1

=(2+

2

)t

1