The position function of a particle in rectilinear motion is given by s(t) = 2t^3 – 21t^2 + 60t + 3 for t ≥ 0. Find the position and acceleration of the particle at the instant the when the particle reverses direction. Include units in your answer.
When the particle reverses direction, this means that there is a sign change in the velocity of the particle or s'(t).
s'(t) = 6t^2 - 42t + 60
Setting it equal to 0 to find critical points;
6t^2 - 42t + 60 = 0
t^2 - 7t + 10 = 0
(t-5)(t-2) = 0
t-5 = 0
t = 5
t-2 = 0
t = 2
The two instances in which the particle reverses direction are t = 2 and t = 5.
To find the acceleration of the particle at these instances, the second derivative of the position function or the first derivative of the velocity function must be taken.
s''(t) = 12t - 42
s''(2) = 12(2) - 42
= 24 - 42
= -18 ft/s^2
s''(5) = 12(5) - 42
= 60 - 42
= 18 ft/s^2
find ds/dt
when it reverses ds/st = 0
find the acceleration d^2s/dt^2 then
To find the position and acceleration of the particle when it reverses direction, we need to find the instantaneous velocity and acceleration at that point.
1. Start by finding the velocity function by taking the derivative of the position function:
v(t) = s'(t) = 6t^2 - 42t + 60
2. The particle changes direction when the velocity is zero. So, we'll set v(t) = 0 and solve for t:
6t^2 - 42t + 60 = 0
3. We can simplify the equation by dividing by 6:
t^2 - 7t + 10 = 0
4. Factor the quadratic equation:
(t - 2)(t - 5) = 0
From this, we find two possible values of t: t = 2 and t = 5.
5. Now, let's find the position and acceleration at each of these instances:
For t = 2:
Plug t = 2 into the position function: s(2) = 2(2)^3 - 21(2)^2 + 60(2) + 3
s(2) = 16 - 84 + 120 + 3
s(2) = 55
Therefore, the position of the particle when it reverses direction at t = 2 is 55 units.
Now, let's find the acceleration at t = 2:
a(t) = v'(t) = 12t - 42
a(2) = 12(2) - 42
a(2) = 24 - 42
a(2) = -18
Therefore, the acceleration of the particle when it reverses direction at t = 2 is -18 units per second squared.
For t = 5:
Plug t = 5 into the position function: s(5) = 2(5)^3 - 21(5)^2 + 60(5) + 3
s(5) = 250 - 525 + 300 + 3
s(5) = 28
Therefore, the position of the particle when it reverses direction at t = 5 is 28 units.
Now, let's find the acceleration at t = 5:
a(t) = v'(t) = 12t - 42
a(5) = 12(5) - 42
a(5) = 60 - 42
a(5) = 18
Therefore, the acceleration of the particle when it reverses direction at t = 5 is 18 units per second squared.
To find the position and acceleration of the particle at the instant it reverses direction, we need to determine the point in time where the particle changes direction. This occurs when the velocity of the particle is equal to zero.
Step 1: Find the derivative of the position function to obtain the velocity function.
Given: s(t) = 2t^3 – 21t^2 + 60t + 3
To find the velocity function, take the derivative of s(t) with respect to t:
v(t) = s'(t) = d/dt (2t^3 – 21t^2 + 60t + 3)
Using the power rule of differentiation, we can differentiate each term:
v(t) = 6t^2 - 42t + 60
Step 2: Set the velocity function equal to zero and solve for t.
To find the point at which the particle reverses direction, we need to find the values of t that make v(t) = 0.
0 = 6t^2 - 42t + 60
This is a quadratic equation, which can be solved by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 6, b = -42, and c = 60. Plugging in these values, we get:
t = (-(-42) ± √((-42)^2 - 4(6)(60))) / (2(6))
Simplifying further:
t = (42 ± √(1764 - 1440)) / 12
t = (42 ± √324) / 12
t = (42 ± 18) / 12
So we have two possible values for t:
t1 = (42 + 18) / 12 = 60/12 = 5
t2 = (42 - 18) / 12 = 24/12 = 2
Therefore, the particle reverses direction at t = 5 and t = 2.
Step 3: Find the position and acceleration at the instants the particle reverses direction.
To find the position at t = 5 and t = 2, substitute these values into the original position function:
s(5) = 2(5)^3 - 21(5)^2 + 60(5) + 3 = 250 - 525 + 300 + 3 = 28
Therefore, the position of the particle at t = 5 is 28 units.
s(2) = 2(2)^3 - 21(2)^2 + 60(2) + 3 = 16 - 84 + 120 + 3 = 55
Therefore, the position of the particle at t = 2 is 55 units.
To find the acceleration at t = 5 and t = 2, we need to find the second derivative of the position function.
Taking the derivative of v(t) = 6t^2 - 42t + 60, we get:
a(t) = v'(t) = d/dt (6t^2 - 42t + 60)
a(t) = 12t - 42
Now we can evaluate the acceleration at the given instants:
a(5) = 12(5) - 42 = 60 - 42 = 18
Therefore, the acceleration of the particle at t = 5 is 18 units per second squared.
a(2) = 12(2) - 42 = 24 - 42 = -18
Therefore, the acceleration of the particle at t = 2 is -18 units per second squared.
In summary:
- The position of the particle at the instant it reverses direction at t = 5 is 28 units.
- The position of the particle at the instant it reverses direction at t = 2 is 55 units.
- The acceleration of the particle at the instant it reverses direction at t = 5 is 18 units per second squared.
- The acceleration of the particle at the instant it reverses direction at t = 2 is -18 units per second squared.