A particle moves along a line so that at time t, where 0<=t<=pi, its position is given by

Question

Create a visually engaging illustration that shows a particle moving along a straight line. The particle's trajectory is influenced by a composite function that combines sinusoidal, quadratic, and constant terms. But don't include any text or specific values associated with the problem, such as 8.13, 1.32, 0.74, -5.19 or 2.55. Instead, portray only the abstract concept of a moving particle under the influence of a mathematical function.

A particle moves along a line so that at time t, where 0<=t<=pi, its position is given by

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

Answer 1

Solve for v(t)

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

Answer 2

T'is actually 3.969. Thank me later

Answer 3

it is 2.55

Answer 4

Well, since options c and b are incorrect, we can rule those out. Now let's find the velocity of the particle when its acceleration is zero.

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!

Answer 5

To find the velocity of the particle when its acceleration is zero, we need to find the derivative of the position function with respect to time and set it equal to zero.

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.

Answer 6

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 .