an object is moving according to the following equation s(t) = 5t - 15t^2 + t^2 s in meter and t in second.
1) identify the times when the distance is maximum and minimum and clearly identify the points.
2)Evaluate the speed at the times in a. Also find the speed and acceleration at t =0,1.5 and 5 seconds.
3)find the time when points of inflexion occur a.
I suspect a typo
why would you have 2 terms in t^2 without adding them together?
Is the last term supposed to be t^3 ?
Clarify before I attempt it.
To find the times when the distance is maximum and minimum, we need to find the critical points of the equation s(t) = 5t - 15t^2 + t^2.
1) To find the maximum and minimum points, we need to find when the derivative of s(t) with respect to t equals zero. The derivative of s(t) is obtained by differentiating each term separately:
s'(t) = 5 - 30t + 2t
Setting this derivative equal to zero, we can solve for t:
5 - 30t + 2t = 0
-28t = -5
t = 5/28 ≈ 0.1786 seconds
This gives us one critical point on the curve. To determine whether it is a maximum or minimum, we can take the second derivative:
s''(t) = -30 + 2
Since the second derivative is positive (2), the critical point corresponds to a minimum.
Therefore, the time when the distance is minimum is approximately t = 0.1786 seconds, and the point is (0.1786, s(0.1786)).
To find the maximum, we can consider the limits as t approaches positive and negative infinity. From the equation, we can see that the term t^2 has a negative coefficient, which means it will eventually dominate as t becomes large or small. Therefore, the maximum point occurs either at t = 0 (starting point) or as t approaches infinity.
The point at t = 0 is (0, s(0)).
For the second part of the problem:
2) To evaluate the speed at the times identified in part 1, we need to find the derivative of s(t):
s'(t) = 5 - 30t + 2t
To find the speed, we take the absolute value of the derivative:
|s'(t)| = |5 - 30t + 2t| = |-28t + 5|
At t = 0.1786 seconds (minimum time), the speed is |s'(0.1786)| = |-28(0.1786) + 5| ≈ 4.9911 m/s
At t = 0 seconds (starting point), the speed is |s'(0)| = |-28(0) + 5| = 5 m/s
To find the speed at t = 1.5 and t = 5 seconds, plug these values into the derivative:
At t = 1.5 seconds:
|s'(1.5)| = |-28(1.5) + 5| ≈ 41 m/s
At t = 5 seconds:
|s'(5)| = |-28(5) + 5| ≈ 135 m/s
To find the acceleration, we need to take the derivative of the speed, which is the second derivative of s(t):
s''(t) = -30 + 2
Substitute the respective values of t:
At t = 0 seconds:
s''(0) = -30 + 2 = -28 m/s^2
At t = 1.5 seconds:
s''(1.5) = -30 + 2 = -28 m/s^2
At t = 5 seconds:
s''(5) = -30 + 2 = -28 m/s^2
3) To find the time when the points of inflection occur, we need to find where the second derivative, s''(t), equals zero. From part 2, we can see that s''(t) is a constant (-28 m/s^2), so it never equals zero. Therefore, there are no points of inflection for this equation.