A particle is moving along the x-axis so that its position at t>=0 is given by s(t)=(t)ln(5t). Find the acceleration of the particle when the velocity is first zero.
s=t*ln(5t)=t(lnt+ln5)
s'=(lnt+ln5)+t(1/t+0)
s"=acceleration=1/t
so find time when s'=0
0=(lnt+ln5)+1
ln5t=-1
5t=e^-1
t= 1/5e
a(1/5e)=1/t=5e
check my work.
To find the acceleration of the particle when the velocity is first zero, we need to find the derivative of the velocity function and then evaluate it when the velocity is zero.
First, let's find the velocity function by taking the derivative of the position function with respect to time:
s(t) = t * ln(5t)
To find the velocity, differentiate s(t) with respect to t:
v(t) = d/dt (t * ln(5t))
Using the product rule for differentiation, the derivative of t * ln(5t) is:
v(t) = 1 * ln(5t) + t * (1/(5t))
Simplifying further:
v(t) = ln(5t) + 1/5
Now we have the velocity function. Next, we want to find when the velocity is first zero. In other words, we need to find the value of t for which v(t) = 0.
Setting v(t) to zero:
0 = ln(5t) + 1/5
Subtracting 1/5 from both sides:
-1/5 = ln(5t)
To eliminate the natural logarithm, we take the exponential of both sides (exponential and natural logarithm are inverse operations):
e^(-1/5) = e^(ln(5t))
Simplifying further:
e^(-1/5) = 5t
Dividing both sides by 5:
(1/5)e^(-1/5) = t
Now we have found the time value when the velocity is zero.
To find the acceleration at this time, we need to find the derivative of the velocity function with respect to time:
a(t) = d/dt (ln(5t) + 1/5)
The derivative of ln(5t) is (1/t), and the derivative of 1/5 is 0. So, the acceleration function is:
a(t) = 1/t
Now we can evaluate the acceleration at the time when the velocity is zero:
a(t) = 1/(1/5)e^(-1/5)
Simplifying further:
a(t) = 5e^(1/5)
Therefore, the acceleration of the particle when the velocity is first zero is 5e^(1/5).