The displacement of a particle moving along the x axis is given by equation x is equals to 2 t cube minus 21 p square + 63 + 6 the possible acceleration of particle when its velocity is zero is
Answers
Given:
The displacement of a particle moving along the x axis is given by equation x is equals to 2 t cube minus 21 p square + 63 + 6
To find:
The possible acceleration of particle when its velocity is zero is
Solution:
From given, we have,
The displacement of a particle moving along the x axis is given by equation x is equals to 2 t cube minus 21 p square + 63 + 6
⇒ x = 2t³ - 21t² +63 + 6
The differentiation of dx/dt represents the velocity.
The differentiation of dv/dt represents the acceleration.
Thus, the differentiation is given by,
dx/dt = 6t² - 42t
The velocity equal to zero implies,
dx/dt = 0
6t² - 42t = 0
6t² = 42t
t = 7 s.
The acceleration is given by,
dv/dt = 12t - 42
substitute the value of t in the above equation.
12t - 42 = 12 (7) - 42 = 84 - 42 = 42 m/s².
Therefore, the possible acceleration of particle when its velocity is zero is 42 m/s².