Question

solve the question plz ​

Answer 1

Explanation:

s= t³-6t²+3t+4 ........(1)

ds/dt= 3t²-12t+3= constant. ..........(2)

(given a=0)

again

calculation of acceleration

d²s/dt²= 6t-12 m/s²= 0 (given)

=> 6t=12=> t= 2sec

this means particle moves with

acceleration upto 2 sec and then gain a

constant velocity.

putting t=2 in velocity eqn

v= 3×2²-12×2+3= 12-24+3= -9m/s

therefore particle moves with a constant velocity

of -9m/s . thus option(4) is correct

Answer 2

Answer:

-9 m/s

Explanation:

differentiate

s=t^3-6t^2+3t+4

you get velocity as

3t^2-12t+3

now as mentioned in question

a=0

but by differentiating velocity you get acceleration

a=6t-12

now

6t-12=0

t=2

now put this value in velocity

v=3(4)-12(2)+3

=12-24+3

= -12+3

= -9

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