The path of a particle moving in a straight line is given by s=t^3-6t^2+9t+4.
a. Find s and a when v=0.
b. Find s and v when a=0.
c. When is s increasing?
d. When is v increasing?
s=t^3-6t^2+9t+4
v = 3t^2 - 12t + 9 = 3(t-1)(t-3)
a = 6t-12 = 6(t-2)
s is increasing when v > 0
v is increasing when a > 0
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a. To find s and a when v = 0, we need to find the velocity function v(t) and acceleration function a(t) first. The velocity is the derivative of the position function s(t), and the acceleration is the derivative of the velocity function.
Given s(t) = t^3 - 6t^2 + 9t + 4, we can find v(t) by taking the derivative:
v(t) = d/dt (t^3 - 6t^2 + 9t + 4)
= 3t^2 - 12t + 9
Now, to find a(t), we take the derivative of v(t):
a(t) = d/dt (3t^2 - 12t + 9)
= 6t - 12
To find s and a when v = 0, we set v(t) = 0 and solve for t:
3t^2 - 12t + 9 = 0
Factoring this equation gives us:
(3t - 3)(t - 3) = 0
So, t = 1 or t = 3.
When t = 1:
s(t) = (1)^3 - 6(1)^2 + 9(1) + 4
= 1 - 6 + 9 + 4
= 8
a(t) = 6(1) - 12
= -6
When t = 3:
s(t) = (3)^3 - 6(3)^2 + 9(3) + 4
= 27 - 54 + 27 + 4
= 4
a(t) = 6(3) - 12
= 6
So, when v = 0, s = 8 or 4, and a = -6 or 6.
b. To find s and v when a = 0, we set a(t) = 0 and solve for t:
6t - 12 = 0
6t = 12
t = 2
When t = 2:
s(t) = (2)^3 - 6(2)^2 + 9(2) + 4
= 8 - 24 + 18 + 4
= 6
v(t) = 3(2)^2 - 12(2) + 9
= 12 - 24 + 9
= -3
So, when a = 0, s = 6 and v = -3.
c. To determine when s is increasing, we need to find the intervals where the derivative of s(t) is positive (greater than 0). The derivative of s(t) is v(t), which we found earlier.
v(t) = 3t^2 - 12t + 9
To solve for v(t) > 0, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we have:
a = 3
b = -12
c = 9
t = (-(-12) ± √((-12)^2 - 4(3)(9))) / (2(3))
= (12 ± √(144 - 108)) / 6
= (12 ± √36) / 6
= (12 ± 6) / 6
Thus,
t1 = (12 + 6) / 6
= 18 / 6
= 3
t2 = (12 - 6) / 6
= 6 / 6
= 1
So, s is increasing for t > 3.
To find the values of s and a when v=0, we need to differentiate the given equation with respect to time (t) to find the velocity function v(t).
The equation for velocity v(t) is obtained by differentiating s(t) with respect to t:
v(t) = ds/dt
So, let's find v(t) by differentiating s(t) = t^3 - 6t^2 + 9t + 4 with respect to t:
ds/dt = 3t^2 - 12t + 9
a. Find s and a when v=0:
To find the values of s and a when v = 0, we need to solve the equation v(t) = 0. So, let's set v(t) = 0 and solve for t:
3t^2 - 12t + 9 = 0
This is a quadratic equation, and we can solve it by factoring or by using the quadratic formula. Factoring gives:
(3t - 3)(t - 3) = 0
From this, we can see that t = 1, or t = 3. These are the values of t when v = 0.
Now, let's find the values of s and a at t = 1 and t = 3.
Plug t = 1 into the equation for s(t):
s(1) = (1)^3 - 6(1)^2 + 9(1) + 4 = 1 - 6 + 9 + 4 = 8
So, when v = 0, s = 8 at t = 1.
Next, find a(t) by differentiating v(t):
a(t) = dv/dt = d^2s/dt^2
Differentiating v(t) = 3t^2 - 12t + 9, we get:
a(t) = 6t - 12
Plug t = 1 into the equation for a(t):
a(1) = 6(1) - 12 = -6
So, when v = 0, a = -6 at t = 1.
Now, repeat the above steps for t = 3:
s(3) = (3)^3 - 6(3)^2 + 9(3) + 4 = 27 - 54 + 27 + 4 = 4
a(3) = 6(3) - 12 = 6
When v = 0, s = 4, and a = 6 at t = 3.
b. Find s and v when a = 0:
To find the values of s and v when a = 0, we need to solve the equation a(t) = 0. So, let's set a(t) = 0 and solve for t:
6t - 12 = 0
Solving this equation gives t = 2.
Now, let's find the values of s and v at t = 2.
Plug t = 2 into the equation for s(t):
s(2) = (2)^3 - 6(2)^2 + 9(2) + 4 = 8 - 24 + 18 + 4 = 6
So, when a = 0, s = 6 at t = 2.
Plug t = 2 into the equation for v(t):
v(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3
So, when a = 0, v = -3 at t = 2.
c. When is s increasing?
To determine when s is increasing, we need to find the values of t where the derivative of s(t) is positive (i.e., ds/dt > 0).
Differentiating s(t) = t^3 - 6t^2 + 9t + 4 with respect to t, we get:
ds/dt = 3t^2 - 12t + 9
We need to find the values of t for which ds/dt > 0. To do this, let's solve the inequality:
3t^2 - 12t + 9 > 0
This can be factored as:
(t - 1)(3t - 9) > 0
From this, we can see that t > 1 or t < 3. So, s is increasing when t is greater than 1 or less than 3.
d. When is v increasing?
To determine when v is increasing, we need to find the values of t where the derivative of v(t) is positive (i.e., dv/dt > 0).
Differentiating v(t) = 3t^2 - 12t + 9 with respect to t, we get:
dv/dt = 6t - 12
We need to find the values of t for which dv/dt > 0. To do this, let's solve the inequality:
6t - 12 > 0
Simplifying this inequality gives:
6t > 12
Dividing both sides by 6, we get:
t > 2
So, v is increasing when t is greater than 2.