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

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Acceleration of a particle moving along a straight line is represented by the relation

$a = 30-4.5 x^2m/s^2.$ The starts with zero initial velocity at

$X = 0.$

written 8.7 years ago by

teamques10 ★ 69k

• modified 4.9 years ago

Determine

(a) The velocity when x = 3m

(b) The position when the velocity is again zero

engineering mechanics

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1 Answer

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written 8.7 years ago by

teamques10 ★ 69k

$a= 30-4.5x^2$

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

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 …

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