Question

A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t) = (t)In(5t). Find the acceleration of the particle when the velocity is first zero.

Answer 1

find the first two derivatives

s(t) = t*ln(5t)

v(t) = s'(t)
a(t) = v'(t)=s''(t)

Answer 2

go ahead take the derivatives

Answer 3

s'(t) = ln5t + 1 ?
s''(t) = 5/5t

Answer 4

right, the functions for position, velocity, and acceleration

s(t)=t*ln(5t)
v(t)=ln(5t)+1
a(t)=1/t


Find a(t) when v(t)=0

Answer 5

where does a(t) come from?

Answer 6

the second derivative a(t)=s''(t), position---velocity---acceleration

Answer 7

it says
Find the acceleration of the particle when the velocity is first zero

When is the velocity first zero? set v(t) = 0 and solve for time t.
v(t)=ln(5t)+1 = 0
ln(5t) =-1
t = e^-1 / 5

Answer 8

The velocity is zero at that time t... so the acceleration at that time is

\[\large a(\frac{ e^{-1} }{ 5 })=\frac{ 1 }{ e^{-1}/5 }=5*e\]

Answer 9

approx... 13.59

Answer 10

Ahh i get it, i forgot to reduce it. Thank you!