Question

A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t) = (t)In(3t). Find the acceleration of the particle when the velocity is first zero.

Answer 1

You must find the derivative twice

Answer 2

would that be 1/x ?

Answer 3

I mean t its 1/t

Answer 4

Yes, that is the acceleration but notice how it says when the velocity is first 0 so what is the velocity?

Answer 5

im not sure

Answer 6

Velocity is the first derivative of displacement, acceleration is the second, so mathematically \[v= \frac{ ds }{ dt }, a = \frac{ d^2s }{ dt^2 }\] so you must have got the first derivative before you got the second, what was it?

Answer 7

(ln)3x +1

Answer 8

Good, \[v = \ln(3t)+1\] now set v = 0 and solve for t, \[0 = \ln(3t)+1\]

Answer 9

I'll start you off, \[\ln(3t)=-1\] any idea how to isolate t?

Answer 10

no

Answer 11

What if we exponentiate both sides

Answer 12

how

Answer 13

\[\huge e^{\ln(3t)}=e^{-1}\]

Answer 14

and then what?

Answer 15

The left side cancels out the logarithm so we're left with \[3t = e^{-1}\]

Answer 16

Finish it off

Answer 17

t=(e^-1)/3

Answer 18

Yes, and use the rule \[x^{-n} = \frac{ 1 }{ x^n }\]

Answer 19

so 1/e divided by 3

Answer 20

\[t = \frac{ 1 }{ 3e }\]

Answer 21

would that be the answer

Answer 22

No, that is your t, now we must plug that in for a that will give us our answer for a at v = 0

Answer 23

1/ 1/3e?

Answer 24

Yes, that simplifies to a(t)=3e

Answer 25

\[\huge \frac{ a }{ \frac{ c }{ d } } = \frac{ ad }{ c }\]