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 when the particle reverses direction. Include units in your answer.
So I've already gotten the first derivative(6t^2-42t+60) and set it to 0. this resulted in t=5 and t=2. I then took the second derivative(12t-42) and plugged in t for acceleration. Where do I go from here?
what you want is the moment when the particle changes direction. That is, when v(t) = 0. t=2 is the first such time, so you just want
s(2) = 55
a(2) = -18
Makes sense, since the velocity is changing from + to -, that the acceleration would be negative.
If you want the values when t=5, just plug that in.
To find the position and acceleration of the particle at the instant when it reverses direction, you have correctly found the critical points of the velocity function (the first derivative). The critical points occur when the velocity is equal to zero, which gives you t = 5 and t = 2.
To determine whether the particle is reversing direction at these points, you can check the sign of the velocity on either side of the critical points. Plug in a value slightly less than 2, such as t = 1, into the velocity function:
v(1) = 6(1)^2 - 42(1) + 60 = 6 - 42 + 60 = 24
Since the velocity is positive (v(1) > 0), the particle is moving in the positive direction between t = 1 and t = 2. Similarly, plug in a value slightly greater than 2, such as t = 3, into the velocity function:
v(3) = 6(3)^2 - 42(3) + 60 = 54 - 126 + 60 = -12
Since the velocity is negative (v(3) < 0), the particle is moving in the negative direction between t = 2 and t = 3.
Therefore, the particle reverses direction at t = 2. Now, let's find the position and acceleration at this instant.
Plug t = 2 into the position function to find the position at t = 2:
s(2) = 2(2)^3 - 21(2)^2 + 60(2) + 3 = 16 - 84 + 120 + 3 = 55
So, the position of the particle at the instant it reverses direction is s = 55 units.
Finally, to find the acceleration at t = 2, plug t = 2 into the acceleration function (the second derivative):
a(2) = 12(2) - 42 = 24 - 42 = -18
Therefore, the acceleration of the particle at the instant it reverses direction is a = -18 units per second squared.
To find the position and acceleration of the particle at the instant when it reverses direction, you have correctly found the values of t when the first derivative of the position function is equal to zero. These values, t = 5 and t = 2, indicate critical points where the particle comes to a stop.
Next, you need to determine whether the particle is changing direction at each of these critical points. To do this, you can check the signs of the first derivative before and after each critical point.
At t = 5, plug in a value slightly less than 5 into the first derivative (6t^2 - 42t + 60). For example, you can use t = 4.9. Evaluating the expression, you get:
f'(4.9) = 6(4.9)^2 - 42(4.9) + 60 = 298.26
Since f'(4.9) > 0, the particle is moving in the positive direction before t = 5. Similarly, plug in a value slightly greater than 5, like t = 5.1:
f'(5.1) = 6(5.1)^2 - 42(5.1) + 60 = -210.66
Since f'(5.1) < 0, the particle is moving in the negative direction after t = 5. Therefore, the particle reverses direction at t = 5.
Repeat the same process for t = 2. Choose a value slightly less than 2, such as t = 1.9:
f'(1.9) = 6(1.9)^2 - 42(1.9) + 60 = -187.02
Since f'(1.9) < 0, the particle is moving in the negative direction before t = 2. Next, choose a value slightly greater than 2, like t = 2.1:
f'(2.1) = 6(2.1)^2 - 42(2.1) + 60 = 213.66
Since f'(2.1) > 0, the particle is moving in the positive direction after t = 2. Therefore, the particle reverses direction at t = 2.
Now that you have confirmed the instances of direction reversal, you can find the position and acceleration of the particle at each of these times.
To find the position at t = 5, substitute t = 5 into the 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 the instant it reverses direction at t = 5 is 28 units.
To find the acceleration at t = 5, substitute t = 5 into the second derivative function (12t - 42):
a(5) = 12(5) - 42
= 60 - 42
= 18
Therefore, the acceleration of the particle at the instant it reverses direction at t = 5 is 18 units per time squared.
Repeat the same process to find the position and acceleration at t = 2.
To find the position at t = 2, substitute t = 2 into the position function:
s(2) = 2(2)^3 - 21(2)^2 + 60(2) + 3
= 16 - 84 + 120 + 3
= 55
Therefore, the position of the particle at the instant it reverses direction at t = 2 is 55 units.
To find the acceleration at t = 2, substitute t = 2 into the second derivative function (12t - 42):
a(2) = 12(2) - 42
= 24 - 42
= -18
Therefore, the acceleration of the particle at the instant it reverses direction at t = 2 is -18 units per time squared.
To summarize:
At instant t = 5:
- The position of the particle is 28 units.
- The acceleration of the particle is 18 units per time squared.
At instant t = 2:
- The position of the particle is 55 units.
- The acceleration of the particle is -18 units per time squared.
Note: Make sure to check your calculations to ensure accuracy.