Question

29.) Position of a particle is given as x = 2t3 - 12t2 + 2, find the acceleration when particle is at rest (1) 48 m/s2 (2) 24 m/s2 (3) 12 m/s2 (4) 36 m/s2 ​

Answer 1

Answer:

The acceleration is 24 m/s^2

option (2) is correct

Explanation:

Given:

$x = 2t^3 - 12t^2 + 2$

then

$velocity=\frac{dx}{dt} = 6t^2-24t$

$acceleration=\frac{d^2x}{dt^2} = 12t-24$

when the particle is at rest, velocity=0

$\implies\:6t^2-24t=0$

$\implies\:6t(t-4)=0$

$\implies\:t=0,4$

when t=0,

$acceleration=12(0)-24=-24\:m/s^2$

when t=4,

$acceleration=12(4)-24=48-24=24\:m/s^2$

Answer 2

Answer:

The acceleration is 24 m/s^2

option (2) is correct

Explanation:

Given:

$x = 2t^3 - 12t^2 + 2$

then

$velocity=\frac{dx}{dt} = 6t^2-24t$

$acceleration=\frac{d^2x}{dt^2} = 12t-24$

when the particle is at rest, velocity=0

$\implies\:6t^2-24t=0$

$\implies\:6t(t-4)=0$

$\implies\:t=0,4$

when t=0,

$acceleration=12(0)-24=-24\:m/s^2$

when t=4,

$acceleration=12(4)-24=48-24=24\:m/s^2$