Question

a particle starts from rest and travels a distance of 6 m along a straight line in two parts, the first part of the journey with constant acceleration of 2m/s^2 and the second part with constant retardation of 4m/s^2.The particle comes to rest at the end of the motion. The total time during which the particle is in motion is?
A. 2s
B.1s
C.3s
D.root3s​

Answer 1

Given that,

Distance = 6 m

Constant acceleration = 2 m/s²

Constant retardation = 4 m/s²

Let the distance of first part is x m and second part is (6-x) m.

We need to calculate the time for first part

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Where, s = distance

a =acceleration

t = time

Put the value into the formula

$x=0+\dfrac{1}{2}\times2\times t^2$

$t=\sqrt{x}$

We need to calculate the velocity at the end of first part

Using equation of motion

$v=u+at$

Put the value into the formula

$v=0+2\sqrt{x}$

$v=2\sqrt{x}$

We need to calculate the velocity at the end of second part

Using equation of motion

$v^2=u^2-2as$

Put the value into the formula

$0=(2\sqrt{x})^2-2\times4\times(6-x)$

$4x-48+8x=0$

$x=\dfrac{48}{12}$

$x=4\ m$

So, The time for first part is

$t=\sqrt{x}$

Put the value of x

$t=\sqrt{4}$

$t=2\ sec$

We need to calculate the time for second part

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$(6-x)=(2\sqrt{x})\times t'+\dfrac{1}{2}\times4\times t'^2$

Put the value of x

$(6-4)=2\times2\times t'+2t'^2$

$2=4t'+2t'^2$

$2t'^2+4t'=2$

$t'^2+2t'-1=0$

$(t'-1)^2=0$

$t'=1\ sec$

We need to calculate the total time during the motion

Using formula for total time

$t''=t+t'$

Put the value into the formula

$t''=2+1$

$t''=3\ sec$

Hence, The total time during the motion is 3 sec.

(C) is correct option.

Answer 2

Answer:

Velocity in the first part,

v1 = u1 + a1 × t1

v1 = 0 + 2t1

v1 = 2t1

Velocity in the second part,

v2 = u2 + a2 × t2

0 = 2t1 - 4t2

2t1 = 4t2

t1/t2 = 2/1

By componendo,

t1 + t2 = 3 sec