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Acceleration of a particle moving along a straight line is represented by the relation a=304.5x2m/s2. The starts with zero initial velocity at X=0.

Determine

(a) The velocity when x = 3m

(b) The position when the velocity is again zero

(c) The position when the velocity is maximum


1 Answer
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a=304.5x2

Substituting a=vdvdxvdvdx=304.5x2vdv=(304.5x2)dx

Integrating both the sides

v2/2=30x4.5x3/3+cv2/2=30x1.5x3+c

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

c is found to be 0

so,

v2/2=30x1.5x3i)v  at  x=3V2=60x3x3=60(3)3(27)=18081=99v=9.95m/sii)x  at  v=00=60x3x360=3x2x=4.47m

iii) x at v maximum

vdv/dx=304.5x2  (given)

When v is maximum, dv/dx=00=304.5x2x=2.58m

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