An object moves along the x-axis, according to x= (pi/2)t + sin(pi t).
a) What is the largest velocity such an object attains?
b) Identify one interval on which the object is accelerating in a positive direction. Explain your reasoning.
c) Name one part of that interval for which the object is speeding up, and explain why you know.
To find the answers to these questions, we need to first understand the given equation and its derivatives. The equation x = (pi/2)t + sin(pi t) represents the position of the object as a function of time, where t is the time in seconds.
a) To find the largest velocity, we need to find the derivative of the position function with respect to time (t). This will give us the velocity function, and the largest velocity will occur when the velocity function is maximized.
The derivative of x with respect to t is obtained by differentiating each term individually. The derivative of (pi/2)t is (pi/2), and the derivative of sin(pi t) is cos(pi t) times the derivative of pi t, which is pi. Thus, the velocity function v(t) = (pi/2) + pi*cos(pi t).
To find the maximum velocity, we need to find the critical points of the velocity function. These occur when the derivative of the velocity function is equal to zero. So, we set dv(t)/dt = 0:
dv(t)/dt = -pi*sin(pi t) = 0
sin(pi t) = 0
This happens when pi t = 0 or pi t = pi. Therefore, the critical points are t = 0 and t = 1. We evaluate the velocity function at these points:
v(0) = (pi/2) + pi*cos(0) = (pi/2) + pi = (2pi + pi)/2 = 3pi/2
v(1) = (pi/2) + pi*cos(pi) = (pi/2) - pi = (-2pi + pi)/2 = -pi/2
Hence, the largest velocity this object attains is |3pi/2| or 3pi/2.
b) To determine the interval in which the object is accelerating in a positive direction, we need to find where the velocity function is positive and its derivative (acceleration function) is also positive. In this case, the velocity function v(t) = (pi/2) + pi*cos(pi t) tells us the object's velocity.
To find when v(t) > 0, we solve:
(pi/2) + pi*cos(pi t) > 0
pi*cos(pi t) > -(pi/2)
cos(pi t) < -1/2
Since the range of cosine is -1 to 1, we need to find where cos(pi t) is less than -1/2. One such interval is the range from t = 0 to t = 1/2. We can verify this by testing values within that interval.
For example, when t = 1/4, cos(pi*1/4) = cos(pi/4) = sqrt(2)/2, which is greater than -1/2. Similarly, when t = 3/4, cos(pi*3/4) = cos(3pi/4) = -sqrt(2)/2, which is less than -1/2. Therefore, the object is accelerating in the positive direction in the interval [0, 1/2].
c) To find a specific part of that interval where the object is speeding up, we need to examine the acceleration of the object. The acceleration is given by the derivative of the velocity function, which is the second derivative of the position function.
By taking the derivative of v(t) = (pi/2) + pi*cos(pi t) with respect to t, we get:
a(t) = -pi^2*sin(pi t)
The sign of the acceleration function will determine whether the object is speeding up (positive acceleration) or slowing down (negative acceleration).
Within the interval [0, 1/2], we observe that when t = 1/4, sin(pi*1/4) = sin(pi/4) = sqrt(2)/2, which is greater than 0. Since the acceleration function a(t) = -pi^2*sin(pi t) is negative, this implies that the object is slowing down in the interval [0, 1/4].
However, when t = 3/8, sin(pi*3/8) = sin(3pi/8) = -sqrt(2)/2, which is less than 0. Since the acceleration function a(t) = -pi^2*sin(pi t) is positive, this means that the object is speeding up in the interval [1/4, 1/2].
Therefore, within the interval [0, 1/2], the object is speeding up in the sub-interval [1/4, 1/2].