Physics, asked by anushkajarange46, 5 hours ago

the position of a particle is given by x=t^3-6t^2+9t. find its position at which velocity becomes zero​

Answers

Answered by vijayalakahmi0901
0

Answer:

A

the body comes to rest firstly at (3−

7

) and then at (3+

7

)

C

the total displacement of the particle in travelling from the first zero of velocity to the second zero of velocity is −74 cm.

D

the particle reverses its velocity at (3−

7

) s and then at (3+

7

) s and has a negative velocity for (3−

7

) < t < (3+

7

)

x

t

=t

3

−9t

2

+6t

V=

dt

d

X

t

=(3t

2

−18t+6)cms

−1

For body to be at rest v=0

⇒t=(3±

7

)s

⇒t

1

=(3−

7

)s and ⇒t

2

=(3+

7

)s

Displacement of particle between t

1

and t

2

is x

t

2

−x

t

1

((3+

7

)

3

−9(3+

7

)

2

+6(3+

7

))−((3−

7

)

3

−9(3−

7

)

2

+6(3−

7

))=−74

v<0for(3−

7

)<t<(3+

7

)

Answered by rainstar720
0

Answer:

4 units

Explanation:

The above attachment contains the explanation.

Attachments:
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