1) A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t) = (t)In(2t). Find the acceleration of the particle when the velocity is first zero.
2) The driver of a car traveling at 50 ft/sec suddenly applies the brakes. The position of the car is s(t) = 50t - 2t^2, t seconds after the driver applies the brakes. How many seconds after the driver applies the brakes does the car come to a stop?
3) The position of a particle on the x-axis at time t, t > 0, is s(t) = e^t with t measured in seconds and s(t) measured in feet. What is the average velocity of the particle for 0 ≤ t ≤ In3?
4) A particle moves along the x-axis so that at any time t, measured in seconds, its position is given by s(t) = sin(t) - 4cos(2t), measured in feet. What is the acceleration of the particle at time t = π seconds?
5) A particle moves with velocity function v(t) = -t^2 + 5t - 3, with v measured in feet per second and t measured in seconds. Find the acceleration of the particle at time t = 3 seconds.
1) To find the acceleration of the particle when the velocity is first zero, we first need to find the velocity function. The velocity is the derivative of the position function.
Given s(t) = t ln(2t), we can find the velocity function v(t) by taking the derivative:
v(t) = d/dt (t ln(2t))
Using the product rule, we can find v(t):
v(t) = ln(2t) + t (d/dt ln(2t))
v(t) = ln(2t) + t (1/(2t))(2)
v(t) = ln(2t) + 1
Now, to find when the velocity is first zero, we need to set v(t) = 0 and solve for t:
ln(2t) + 1 = 0
Subtracting 1 from both sides:
ln(2t) = -1
Taking the exponential of both sides:
e^(ln(2t)) = e^(-1)
Simplifying:
2t = 1/e
Dividing both sides by 2:
t = 1/(2e)
To find the acceleration, we need to take the derivative of the velocity function:
a(t) = d/dt (ln(2t) + 1)
Using the chain rule and derivative of natural logarithm, we can find a(t):
a(t) = (1/t)(2) = 2/t
Substituting t = 1/(2e):
a(t) = 2/(1/(2e)) = 4e
Therefore, the acceleration of the particle when the velocity is first zero is 4e.
2) To find how many seconds after the driver applies the brakes does the car come to a stop, we need to set the velocity function v(t) equal to zero and solve for t.
Given v(t) = 50 - 4t, we set v(t) = 0 and solve for t:
50 - 4t = 0
Subtracting 50 from both sides:
-4t = -50
Dividing by -4:
t = 12.5
Therefore, it takes 12.5 seconds for the car to come to a stop after the driver applies the brakes.
3) To find the average velocity of the particle for 0 ≤ t ≤ ln(3), we need to calculate the displacement and divide it by the time interval.
The displacement is given by subtracting the initial position from the final position:
s(f) - s(i) = s(ln(3)) - s(0) = e^(ln(3)) - e^(0) = 3 - 1 = 2
The time interval is ln(3) - 0 = ln(3).
Therefore, the average velocity is displacement / time interval:
average velocity = 2 / ln(3)
4) To find the acceleration of the particle at time t = π seconds, we need to take the second derivative of the position function.
Given s(t) = sin(t) - 4cos(2t), taking the second derivative will give us the acceleration function.
The first derivative:
v(t) = d/dt (sin(t) - 4cos(2t))
v(t) = cos(t) + 8sin(2t)
The second derivative:
a(t) = d/dt (cos(t) + 8sin(2t))
a(t) = -sin(t) + 16cos(2t)
Substituting t = π:
a(π) = -sin(π) + 16cos(2π)
a(π) = 0 + 16(1)
a(π) = 16
Therefore, the acceleration of the particle at time t = π seconds is 16.
5) To find the acceleration of the particle at time t = 3 seconds, we need to take the derivative of the velocity function.
Given v(t) = -t^2 + 5t - 3, we can find the acceleration function a(t) by taking the derivative.
a(t) = d/dt (-t^2 + 5t - 3)
a(t) = -2t + 5
Substituting t = 3:
a(3) = -2(3) + 5
a(3) = -6 + 5
a(3) = -1
Therefore, the acceleration of the particle at time t = 3 seconds is -1.
1) To find the acceleration of the particle when the velocity is first zero, we need to differentiate the velocity function with respect to time and then find the value of t when the velocity is zero.
First, let's find the velocity function by differentiating the position function with respect to time.
s(t) = t * ln(2t)
Using the product rule of differentiation, we can differentiate the function as follows:
v(t) = t * d(ln(2t))/dt + ln(2t) * d(t)/dt
= t * (1/(2t))*2 + ln(2t) * 1
= 1 + ln(2t)
Now, to find when the velocity is zero, we set v(t) = 0 and solve for t:
0 = 1 + ln(2t)
Subtracting 1 from both sides:
ln(2t) = -1
Taking the exponential of both sides:
2t = exp(-1)
Simplifying:
t = exp(-1)/2
Now that we have the value of t, we can find the acceleration by differentiating the velocity function with respect to time:
a(t) = d(v(t))/dt
Since v(t) = 1 + ln(2t), we can differentiate it using the chain rule:
a(t) = d(1 + ln(2t))/dt
= 0 + d(ln(2t))/dt
= 1/(2t)
Therefore, the acceleration of the particle when the velocity is first zero is a(t) = 1/(2t).
2) To find the time it takes for the car to stop, we need to determine when its velocity becomes zero.
The velocity function is given as v(t) = 50 - 4t.
To find when the velocity is zero, we set v(t) = 0 and solve for t:
0 = 50 - 4t
Simplifying:
4t = 50
t = 50/4
t = 12.5
Therefore, the car comes to a stop after 12.5 seconds.
3) To find the average velocity of the particle for the given time interval, we need to calculate the displacement and divide it by the time elapsed.
The position function is given as s(t) = e^t.
To find the displacement from t = 0 to t = ln(3), we subtract the value of the position function at t = 0 from the value at t = ln(3):
s(ln(3)) - s(0) = e^ln(3) - e^0
= 3 - 1
= 2
The time elapsed is ln(3) - 0 = ln(3).
Therefore, the average velocity of the particle for 0 ≤ t ≤ ln(3) is 2/ln(3) ft/sec.
4) To find the acceleration of the particle at time t = π seconds, we need to differentiate the velocity function with respect to time, and then substitute t = π into the resulting expression.
The position function is given as s(t) = sin(t) - 4cos(2t).
To find the velocity function, we differentiate the position function with respect to time:
v(t) = d(s(t))/dt
= d(sin(t))/dt - d(4cos(2t))/dt
= cos(t) + 8sin(2t)
Now, let's find the acceleration by differentiating the velocity function with respect to time:
a(t) = d(v(t))/dt
= d(cos(t) + 8sin(2t))/dt
= -sin(t) + 16cos(2t)
Substituting t = π into the acceleration function:
a(π) = -sin(π) + 16cos(2π)
= 0 - 16
= -16
Therefore, the acceleration of the particle at time t = π seconds is -16 ft/sec^2.
5) To find the acceleration of the particle at time t = 3 seconds, we simply substitute t = 3 into the acceleration function.
The velocity function is given as v(t) = -t^2 + 5t - 3.
To find the acceleration, we differentiate the velocity function with respect to time:
a(t) = d(v(t))/dt
= d(-t^2 + 5t - 3)/dt
= -2t + 5
Substituting t = 3 into the acceleration function:
a(3) = -2(3) + 5
= -6 + 5
= -1
Therefore, the acceleration of the particle at time t = 3 seconds is -1 ft/sec^2.
I am not going to just do these for you. Did you try? Where did you get stuck?
Plug and chug, for example:
ds/dt = t* d/dt(ln 2t) + ln 2t
= t * (1/2t)2 + ln 2 t
= 1 + ln 2t
when is that zero?
ln 2t = -1
e^ln 2t = 2t =1/e
t = 1/(2e)
a = d^2s/dt^2 = d/dt(ds/dt)
= 0 + d/dt(ln2t) = (1/2t)2 = 1/t
so when t = 1/2e
that acceleration is 2e