velocity??

Question

the formula x(t)=lnt + (t^2/18) + 1 gives the position of an object moving along the x-axis during the time interval 1 ≤t ≤ 5. At the instant when the acceleration of the object is zero, the velocity is
a)0
b)1/3
c)2/3
d)1
e)undefined

Answer 1

velocity: v = dx/dt = 1/t + t/9
acceleration: a = dv/dt = -1/(t^2) + 1/9
When acceleration is zero,
a = 0 = -1/(t^2) + 1/9
i.e. t = 3 [since time cannot be negative]
So velocity at time t = 3
v = 1/3 + 3/9 = 2/3

c) is correct

Answer 2

For the given curve , the acceleration , which is the second derivative of the equation is :

d/dt(1/t + 2t/18) i.e. -1/t^2 + 1/9.

When this is zero , we get -9 + t^2 = 0 , which means t^2 = 9.

Which means the instant at which the acceleration is zero is t = 3.

At this instant , the velocity , which is the first derivative , is :

1/t + t/9 = 1/3 + 1/3 = 2/3

Hence the answer is (c) 2/3.

Answer 3

what is Int ( constant? )

take the differentiate the equation twice to get the acceleration equation. Equate that to zero. Substitute the results into the velocity equation to solve it

Answer 4

v = dx/dt = (1/t) +(2t/18)

v = (1/t) +(t/9)

a = dv/dt = -1/t^2 +(1/9)

a = 0 , then -(1/t^2) +(1/9) = 0
(1/t^2) = 1/9
9 = t^2

t = -3 ( time cant be negative )and t = 3

v = (1/3) +(6/18) = (1/3) +(1/3) = 2/3

Answer 5

Ok i will tell you how to do it. Acceleration is the rate of change of velocity ( or the DERIVATIVE of velocity). So take the derivative of the equation, and equate it to 0. So you know where the accerleation equation (deriv. of velocity) equals zero. Then, just plug the value that makes the acceleration 0 into the velocity equation.

Answer 6

1/3

Answer 7

dont know

Answer 8

v = dx / dt = 1 / t + ( 1 / 9 ) t
dv / dt = - 1 / t ² + 1 / 9 = 0
1 / t ² = 1 / 9
t = 3
v = 1 / 3 + 1 / 3
v = 2 / 3

OPTION c)

Answer 9

Thanks for all the answers