the position of a body at time t sec is s=t^3 - 6t^2 + 9t m. Find the body's acceleration each time the velocity is zero.
velocity is ds/dt = 3t^2 - 12t + 9
we need t when velocity is zero, so
3t^2 - 12y + 9 = 0
t^2 - 4y + 3 = 0
(t-1)(t-3) = 0
t = 1 , t = 3
a = dv/dt = 6t - 12
when t = 1, (v=0), a = 6-12 = -6 m/s^2
when t = 3 , a = 18-12 = 6 m/s^2
Well, it seems like the body is having some trouble getting around, doesn't it? Let's put on our detective hats and figure this out.
First, let's find the velocity of the body by taking the derivative of the position function. The velocity function is given by v(t) = 3t^2 - 12t + 9 m/s.
To find when the velocity is zero, we set v(t) equal to zero and solve for t. So 3t^2 - 12t + 9 = 0.
Now, let me tell you a little secret about quadratic equations. They can be solved using the quadratic formula, which goes something like this: t = (-b ± √(b^2 - 4ac))/(2a).
So plugging in the coefficients of our equation, a = 3, b = -12, and c = 9, we can find the values of t when the velocity is zero.
Using the formula, we get t = (12 ± √(144 - 108))/6. Simplifying that, we find t = (12 ± √36)/6.
Now let's do a little math to find the values of t. We have t = (12 ± 6)/6, which gives us t = 3 or t = 2.
So, when the velocity is zero, the body's acceleration is given by the second derivative of the position function. Taking the derivative of the velocity function, we get a(t) = 6t - 12 m/s².
Therefore, when t = 2 or t = 3, the body's acceleration is a(2) = -18 m/s² or a(3) = 6 m/s², respectively.
I hope that shines a light on the situation for you!
To find the body's acceleration at the times when the velocity is zero, we first need to calculate the velocity function. Velocity is the derivative of displacement with respect to time.
Given that the position function of the body is:
s(t) = t^3 - 6t^2 + 9t
To find the velocity function, we need to differentiate the position function with respect to time:
v(t) = d/dt[s(t)]
Taking the derivative of each term, we have:
v(t) = d/dt[t^3 - 6t^2 + 9t]
v(t) = 3t^2 - 12t + 9
Now we can determine the times when the velocity is equal to zero by solving the equation:
3t^2 - 12t + 9 = 0
We can factor the above equation as follows:
3(t^2 - 4t + 3) = 0
3(t - 1)(t - 3) = 0
Therefore, the possible solutions are t = 1 and t = 3.
Now, to find the acceleration at these times, we differentiate the velocity function with respect to time.
a(t) = d/dt[v(t)]
Taking the derivative of each term, we have:
a(t) = d/dt[3t^2 - 12t + 9]
a(t) = 6t - 12
Substituting t = 1 and t = 3 into the acceleration function:
a(1) = 6(1) - 12 = -6 m/s^2
a(3) = 6(3) - 12 = 6 m/s^2
Therefore, the body's acceleration each time the velocity is zero is -6 m/s^2 at t = 1 sec, and 6 m/s^2 at t = 3 sec.
To find the body's acceleration when the velocity is zero, we need to follow a few steps.
Step 1: Find the expression for velocity by taking the derivative of the position function with respect to time.
The position function is given as s = t^3 - 6t^2 + 9t.
To find the velocity, we take the derivative of the position function with respect to time (t):
v = ds/dt
v = d/dt(t^3 - 6t^2 + 9t)
v = 3t^2 - 12t + 9
Step 2: Set the velocity expression equal to zero and solve for t to find when the velocity is zero.
Setting v = 0, we have:
0 = 3t^2 - 12t + 9
This is a quadratic equation. We can solve it using factoring, completing the square, or using the quadratic formula.
Step 3: Solve the quadratic equation to find the values of t when the velocity is zero.
Using the quadratic formula: t = (-b ± √(b^2-4ac)) / (2a)
In our case, a = 3, b = -12, and c = 9.
Plugging these values into the formula, we get:
t = (-(-12) ± √((-12)^2 - 4(3)(9))) / (2(3))
= (12 ± √(144 - 108)) / 6
= (12 ± √36) / 6
Simplifying further, we get:
t = (12 ± 6) / 6
This gives us two possible values for t:
t1 = (12 + 6) / 6 = 3
t2 = (12 - 6) / 6 = 1
Step 4: Substitute the values of t back into the acceleration formula to find the body's acceleration.
The acceleration of the body can be found by taking the derivative of the velocity function:
a = dv/dt
a = d/dt(3t^2 - 12t + 9)
a = 6t - 12
Substituting t = 1 into the acceleration formula:
a = 6(1) - 12 = -6 m/s^2
Substituting t = 3 into the acceleration formula:
a = 6(3) - 12 = 6 m/s^2
Therefore, when the velocity is zero, the body's acceleration is -6 m/s^2 and 6 m/s^2.