Question: 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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Determine

(a) The velocity when x = 3m

(b) The position when the velocity is again zero

(c) The position when the velocity is maximum


Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 04

Years : DEC 2014

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modified 13 months ago by gravatar for Yashbeer Yashbeer170 written 3.4 years ago by gravatar for Aksh_31 Aksh_31680
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$$a= 30-4.5x^2 $$

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 \\ i) v \space \space at\space \space x=3 \\ V^2= 60x-3x^3 \\ = 60(3)-3(27) \\ =180-81=99 \\ v= 9.95 m/s \\ ii) x\space \space at\space \space v=0 \\ 0= 60x-3x^3 \\ 60=3x^2 \\ x=4.47m $

iii) x at v maximum

$vdv/dx = 30-4.5x2 \space \space (given) $

When v is maximum, $dv/dx=0\\ 0 = 30 - 4.5x^2 \\ x= 2.58m$

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written 3.4 years ago by gravatar for Aksh_31 Aksh_31680
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