Question

The position of a particle moving in the xy plane at any time t is given by (3t ​​2 - 6t , t 2 - 2t)m. Select the correct statement about the moving particle from the following: its acceleration is never zero particle started from origin (0,0) particle was at rest at t= 1s at t= 2s velocity and acceleration is parallel

Answer 1

x = 3 t² - 6 t   meters          y = t² - 2 t   m
x = 3 t (t - 2)   m                  y = t (t - 2)  m
x = 3 y
Angle made by displacement with x axis = Tan⁻¹ (1/3) 

v_x = 6 t - 6    m/s              v_y = 2 t - 2      m/s
v_x = 3 v_y
Angle made by velocity with x axis :  Tan⁻¹ (1/3)

a_x  = 6   m/s^2                   a_y = 2   m/s^2
a_x = 3  a_y
angle made by acceleration with x axis = Tan⁻¹ 1/3

a (x,y) = √(36+4)  = 2√10  m/sec^2       acceleration is never zero.

v (x,y,t) = √(v²_x +v²_y) = √10 v_y = 2 √10 (t - 1)   m/s
velocity is zero  at   t = 1 sec...

v(t=2 sec) = 2√10 m/s

At t = 2sec, the velocity and acceleration vectors are parallel., as angles made by them with x axis are same

Answer 2

The acceleration and acceleration are parallel at 2 seconds.

Explanation

Given data:

X = 3 t^2 - 6 t   m  ,Y = t^2 - 2 t   m

X = 3 t (t - 2)   m  , Y = t (t - 2)  m

X = 3 y

θ along x axis = tan^-1 (1 / 3) 

Now velocity long X and Y axis.

V (x)  = 6 t - 6   m/s  

V (y) = 2 t - 2    m/s

V (x) = 3 V (y)

θ along x axis :  tan^-1 (1 / 3)

a along (x)  axis = 6   m/s^2  , a long (y) axis = 2   m/s^2

a (x) = 3  a (y)

θ made by acceleration with x axis = tan⁻¹ (1 / 3)

a (x,y) = √(36+4)  = 2√10  m/sec^2    

V (x,y,t) = √[(v² (x) +v^2(y)] = √10 v(y)

V (x,y,t)  = 2 √10 (t - 1)   m/s

Velocity is "0" at 1 sec.

Velocity when (t = 2 sec) = 2√10 m/s

The velocity and acceleration vectors are parallel to each other at t = 2 sec.