
Patrick B. answered 01/01/20
Math and computer tutor/teacher
y = t ln (5t)
y ' = t * 5/(5t) + ln(5t) = 1 + ln(5t)
the velocity is zero when 1 + ln(5t) = 0
ln(5t) = -1
5t = 1/e
t = 1/(5e)
second derivative is y'' = 5/(5t) = 1/t
the acceleration at that time is then 1/(1/(5e)) = 5e