Question

The displacement (x) of a particle moving along
x-axis varies with time (t) as x = (2t^2-3t^3) m,
where t is in seconds, then
(1) The particle returns to its initial position after
3/2
4
* (2) The particle comes to rest at t= 4/9
$(3) Initial velocity of particle is zero
(4) Maximum velocity in positive x-direction is
4/9m/s
​

Answer 1

Answer:

Step by Step Experience

Answer 2

Given info : The displacement (x) of a particle moving along

x-axis varies with time (t) as x = (2t² -3t³ ) m.

To check the correct statement(s) :

1. The particle returns to its initial position after  3/2  
2. The particle comes to rest at t= 4/9
3. Initial velocity of particle is zero
4. Maximum velocity in positive x-direction is  4/9m/s

solution : x = (2t² - 3t³ ) m

at t = 0 , initial position , x₀ = 2(0)² - 3(0)³ = 0

we see, at x = 0, 2t² - 3t³ = 0 ⇒ t = 0, 2/3

therefore after 2/3 sec , the particle returns to its initial position.

differentiate x with respect to t, dx/dt = 4t - 9t²

⇒ velocity of the particle will be zero if dx/dt = 0

∵ 4t - 9t² = 0

⇒ t = 0 , 4/9

therefore the particle comes to rest at t = 4/9 sec.

and initial velocity of the particle, $\frac{dx}{dt}\left|_{t=0}$ = 0 .

dx/dt = 4t - 9t² , differentiate it with respect to t,

⇒ d²x/dt² = 4 - 18t = 0  ⇒ t = 2/9

now the maximum velocity , $\frac{dx}{dt}\left|_{t=\frac{2}{9}}$ = $4\left(\frac{2}{9}\right)-9\left(\frac{2}{9}\right)^2$= 4/9 m/s

therefore the maximum velocity in positive x direction is 4/9 m/s.

therefore the correct options are (2) , (3) and (4).