The position of a particle in rectilinear motion is given by s(t) = tln(1/t) for t>0, where t is in seconds.
Find the time (t>0) at which the velocity of the particle is 0, and find the particle's acceleration at that time as well. Use exact values.
s(t) = tln(1/t)
= t(ln1 - lnt)
= t(0 - lnt)
= -t lnt
v(t) = -t(1/t) - lnt
= -1 - lnt
a(t) = 0 - 1/t = -1/t
we want v(t) = 0
-1- lnt = 0
lnt = -1
t = 1/e
a(1/e) = -1/(1/e) = -e
Oh, I see you're looking for some mathematical amusement! Well, let's find the time at which the velocity of the particle is 0 and the particle's acceleration at that time.
The velocity of the particle can be found by taking the derivative of the position function with respect to time.
So, let's find the derivative of s(t) = tln(1/t).
Using the product rule and chain rule, we have:
s'(t) = (t)(d/dt)[ln(1/t)] + ln(1/t)(d/dt)(t)
= (t)(-1/t^2) + ln(1/t)(1)
= -1/t + ln(1/t)
Now, to find the time at which the velocity is 0, we set s'(t) = 0 and solve for t:
-1/t + ln(1/t) = 0
Well, I'm afraid I can't give you the exact value for t, but you can solve this equation numerically to find the value of t that makes the velocity 0.
Now, to find the particle's acceleration at that time, we need to take the derivative of the velocity function.
So, let's find the derivative of s'(t) = -1/t + ln(1/t).
Using the chain rule, we have:
s''(t) = (d/dt)(-1/t) + (d/dt)(ln(1/t))
= 1/t^2 - (1/t)(d/dt)(1/t)
= 1/t^2 - (1/t)(-1/t^2)
= 1/t^2 + 1/t^3
Again, I'm sorry I can't provide you with the exact value of the acceleration at that time, as it depends on the specific value of t you find. Nevertheless, you can calculate it numerically once you determine the time at which the velocity is 0.
Remember, math can be fun when you add a little humor!
To find the time at which the velocity of the particle is 0, we need to find the value of t for which the derivative of the position function, s(t), equals zero.
The velocity function can be found by taking the derivative of the position function with respect to t:
v(t) = s'(t)
To find v(t), we need to find the derivative of tln(1/t) with respect to t.
Using the chain rule, the derivative of ln(1/t) with respect to t is:
d/dt [ln(1/t)] = (1/1/t) * (-1/t^2) = -t
Hence, the velocity function becomes:
v(t) = -t
To find the time t at which v(t) = 0, we set -t = 0 and solve for t:
-t = 0
t = 0
Therefore, the time at which the velocity of the particle is 0 is t = 0.
To find the particle's acceleration at that time, we need to find the derivative of the velocity function with respect to t:
a(t) = v'(t)
To find v'(t), we differentiate -t with respect to t:
d/dt [-t] = -1
Hence, the acceleration function becomes:
a(t) = -1
Therefore, at t = 0, the particle's acceleration is -1.
To find the time at which the velocity of the particle is 0, we need to find the value of t when the derivative of the position function, s(t), equals zero. The derivative of s(t), denoted as v(t), represents the velocity of the particle.
First, let's find v(t) by taking the derivative of s(t):
v(t) = d/dt (t * ln(1/t))
To differentiate this expression, we can use the product rule. The product rule states that if we have two functions u(t) and v(t), then the derivative of their product is given by:
d/dt (u(t) * v(t)) = u'(t) * v(t) + u(t) * v'(t)
Let's use the product rule to find v(t):
u(t) = t
v(t) = ln(1/t)
Now, let's find the derivatives of u(t) and v(t):
u'(t) = 1
v'(t) = d/dt(ln(1/t))
To differentiate ln(1/t), we can use the chain rule, which states that if we have a composite function f(g(t)), then the derivative of f(g(t)) is given by:
d/dt(f(g(t))) = f'(g(t)) * g'(t)
In our case, f(x) = ln(x), and g(t) = 1/t. Therefore:
f'(x) = 1/x
g'(t) = d/dt(1/t) = -1/t^2
Now, let's differentiate ln(1/t) using the chain rule:
d/dt(ln(1/t)) = (1/(1/t)) * (-1/t^2) = -1/t
Finally, let's substitute these derivatives into the product rule we derived earlier:
v'(t) = u'(t) * v(t) + u(t) * v'(t)
v'(t) = 1 * ln(1/t) + t * (-1/t)
v'(t) = ln(1/t) - 1
Now, let's set the derivative v'(t) equal to zero and solve for t:
ln(1/t) - 1 = 0
To get rid of the natural logarithm, we can exponentiate both sides of the equation with base e:
e^(ln(1/t) - 1) = e^0
e^(ln(1/t)) * e^(-1) = 1
(1/t) * e^(-1) = 1
Now, let's solve for t by multiplying both sides of the equation by t:
e^(-1) = t
Therefore, at t = e^(-1) (approximately 0.368), the velocity of the particle is 0.
Next, let's find the acceleration of the particle at t = e^(-1). The acceleration, denoted as a(t), can be found by taking the derivative of the velocity function, v(t):
a(t) = d/dt(v(t))
We already found v(t) earlier:
v(t) = ln(1/t) - 1
To find a(t), we need to differentiate v(t):
a(t) = d/dt(ln(1/t) - 1)
To differentiate -1, we get 0, and to differentiate ln(1/t), we can use the chain rule, similar to what we did earlier:
a(t) = d/dt(ln(1/t)) - d/dt(1)
a(t) = -1/t - 0
a(t) = -1/t
Therefore, at t = e^(-1) (approximately 0.368), the particle's acceleration is -1/(e^(-1)).