Acceleration of a particle moving along a straight line is represented by the relation $a = 30-4.5x^2m/s^2$ . The starts with zero initial velocity at $ X = 0. $

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written 8.9 years ago by

teamques10 ★ 70k

modified 3.8 years ago by

yashbeer ★ 11k

Determine

(a) the velocity when x = 3m

(b) the position when the velocity is again zero

engineering mechanics

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written 8.9 years ago by

teamques10 ★ 70k

$$a= 30-4.5x^2$$

Substituting $a = v dv/dx \\ vdv/dx = 30-4.5x^2 \\ v dv = (30-4.5x^2)dx$

Integrating both the sides

$v^2/2=30x-4.5x^3/3 + c \\ v^2/2=30x-1.5x^3+ c $

Put v= 0 at x=0 (given) to get value of c.

c is found to be 0

so,

$v^2/2=30x-1.5x^3 \\ i) v\space at\space …

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