Your function is odd, I read it as
Cosine ( t-((t^2/2) ) + 10
check that
Take the derivative, that is velocity
Take the derivative again, and set to zero, solve for t. Put that t into velocity .
s(t)= -4cost-((t^2)/2)+10 What is the velocity of the particle when its acceleration is zero?
a. 8.13
b. 1.32
c. 0.74
d. -5.19
e. 2.55
I know that c and b are incorrect
Please help me
Cosine ( t-((t^2/2) ) + 10
check that
Take the derivative, that is velocity
Take the derivative again, and set to zero, solve for t. Put that t into velocity .
V(t) =4sin(t)-t
Solve for a(t)
a(t)=4cos(t)-1
Find when a(t)=0
t=1.32
Plug t into v(t)
ANSWER=2.55
To find the acceleration, we need to find the second derivative of the position function. Let's do that:
s(t) = -4cos(t) - (t^2)/2 + 10
Taking the derivative with respect to time:
s'(t) = 4sin(t) - t
Now let's find the second derivative:
s''(t) = 4cos(t) - 1
To find when the acceleration is zero, we set s''(t) = 0 and solve for t:
4cos(t) - 1 = 0
cos(t) = 1/4
Since cosine is positive in the first and fourth quadrants, we can say:
t = arccos(1/4) or t = -arccos(1/4)
Now, let's substitute these values of t back into the velocity function (s'(t)) to find the velocity when the acceleration is zero:
s'(arccos(1/4)) = 4sin(arccos(1/4)) - arccos(1/4)
s'(-arccos(1/4)) = 4sin(-arccos(1/4)) - (-arccos(1/4))
Calculating these values, we get:
s'(arccos(1/4)) ≈ 2.55
s'(-arccos(1/4)) ≈ -5.19
So, the velocity of the particle when its acceleration is zero is approximately 2.55 or -5.19.
Therefore, the correct answer is d. -5.19.
I hope that puts a smile on your face!
Given:
s(t) = -4cos(t) - (t^2)/2 + 10
First, let's find the derivative of s(t) with respect to t:
s'(t) = -4(-sin(t)) - (2t)/2
s'(t) = 4sin(t) - t
Now, we need to set s'(t) equal to zero:
4sin(t) - t = 0
To solve this equation, we can use numerical methods or approximate values to find when the expression equals zero.
In the given range 0 <= t <= pi, the equation 4sin(t) - t = 0 has a root close to t = 1.32.
Now, let's find the velocity of the particle at t = 1.32.
To find the velocity, we substitute t = 1.32 into the derivative function s'(t):
s'(1.32) = 4sin(1.32) - 1.32
Evaluating this expression, we find that the velocity is approximately -1.58.
Since none of the provided answer choices match -1.58, there may be an error or discrepancy in the given choices. It's possible that the correct answer is not included in the options provided.