Question

The position of a particle varies with time
according to the relation x = 3t2 –t³ m. The
maximum positive velocity of the particle (in
m/s) and distance travelled (in m) by the particle
upto 4 sec. respectively are :-

1)3,16
2)3,24
3)3,32
4)2,32



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Answer 1

Answer :

Explanation :

As we have the eq^n

X=3t^2-t^3

Therefore, by taking derivetion on both side

X=6t-3t^2

As, 6t=3t^2

6=3t

t=2sec.

Substitute in eq^n,

V=dx/dt

V=3(2)^2-(2)^3/2

=12-8/2

=6/2

V=3m/s

Distance, upto 4 sec is

X=3t^2-t^3

=3(4)^2-(4)^3

=48-64

=(-16)

As distance never be (- ve)

Therefore, x=16m

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