Question

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⤴⤴Refer the attachment⤴⤴

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Answer 1

Answer:

The time required to come to rest is 1 second

Explanation:

The displacement is given as:

x=10+6t-3t^2

Differentiate it wrt to time (t) we get,

dx/dt=6-6t

v=6-6t

At rest velocity =0

0=6-6t

t=1

hence the time required is 1 second

Answer 2

THE GIVEN EQUATION IS 》》

$10 + 6t - {3t}^{2}$

NOW -☆differentiation☆

$= > 0 + 6 - 6t \\ \\ > = > 6 - 6t$

THE TIME TAKEN BY PARTICLE TO BE IN REST.

$= > 6 -6t = 0 \\ \\ = > t = 1$

AFTER ONE SECOND THE PARTICLE WILL BE AT REST.