A particle moves along the x-axis so that it’s position s in meters at any time t>0 in seconds is given by the function s(t)=2t+2sin((pi/2)t).
A. Find the velocity of the particle, and then determine the velocity at times t=1 and t=2.
B. Is the particle speeding up or slowing down at t=1.9 seconds? Justify your answer.
C. Find the time t on the interval [0,3] at which the particle is farthest to the right. Justify your answer.
whats c
To find the velocity of the particle, we need to differentiate the position function with respect to time.
Given: s(t) = 2t + 2sin((pi/2)t)
A. Velocity:
Taking the derivative of s(t) with respect to t:
s'(t) = d/dt [2t + 2sin((pi/2)t)]
= 2 + 2cos((pi/2)t) * (pi/2)
= 2 + (pi/2)cos((pi/2)t)
Therefore, the velocity of the particle is v(t) = 2 + (pi/2)cos((pi/2)t).
To find the velocity at t=1, substitute t=1 into the velocity function:
v(1) = 2 + (pi/2)cos((pi/2)*1)
= 2 + (pi/2)cos(pi/2)
= 2 + (pi/2)(0)
= 2
So, the velocity of the particle at t=1 is 2 meters per second.
Similarly, to find the velocity at t=2, substitute t=2 into the velocity function:
v(2) = 2 + (pi/2)cos((pi/2)*2)
= 2 + (pi/2)cos(pi)
= 2 - (pi/2)(-1)
= 2 + (pi/2)
= (4 + pi)/2
= 2 + pi/2
So, the velocity of the particle at t=2 is (2 + pi/2) meters per second.
B. To determine whether the particle is speeding up or slowing down at t=1.9 seconds, we need to examine the sign of the velocity. If the velocity is positive, the particle is speeding up; if the velocity is negative, the particle is slowing down.
Substituting t=1.9 into the velocity function:
v(1.9) = 2 + (pi/2)cos((pi/2)*1.9)
To determine the sign of v(1.9), we need to evaluate cos((pi/2)*1.9). Since cosine is positive in the first and fourth quadrants, and (pi/2)*1.9 falls in the second quadrant, we have cos((pi/2)*1.9) < 0.
Since v(1.9) = 2 + (pi/2)cos((pi/2)*1.9), and cos((pi/2)*1.9) < 0, the velocity at t=1.9 is negative. Therefore, the particle is slowing down at t=1.9 seconds.
C. To find the time t on the interval [0,3] at which the particle is farthest to the right, we need to find the maximum value of the position function s(t).
The position function s(t) = 2t + 2sin((pi/2)t) can be maximized by finding the critical points or endpoints of the interval [0,3].
First, we find the critical points by finding where the derivative of s(t) equals zero:
v(t) = 2 + (pi/2)cos((pi/2)t)
v'(t) = -(pi^2/4)sin((pi/2)t)
Setting v'(t) = 0, we solve for t:
-(pi^2/4)sin((pi/2)t) = 0
sin((pi/2)t) = 0
The possible critical points are:
t = 0, t = 2/π, t = 4/π, t = 6/π, t = 8/π, ...
Checking the endpoints of the interval [0,3]:
s(0) = 2*0 + 2sin((pi/2)*0) = 0
s(3) = 2*3 + 2sin((pi/2)*3) = 6 + 2sin((3pi/2)) = 6 - 2 = 4
Comparing the values of s(0) and s(3), we can see that s(3) > s(0).
Therefore, the particle is farthest to the right at t = 3 seconds on the interval [0,3].
To find the velocity of the particle, we need to take the derivative of the position function, s(t). The velocity function, v(t), is obtained by taking the derivative of s(t) with respect to time, t.
A. Calculating the velocity of the particle:
We have s(t) = 2t + 2sin((π/2)t).
To find the velocity function v(t):
v(t) = d(s(t))/dt.
Taking the derivative of s(t) with respect to t:
v(t) = d/dt (2t + 2sin((π/2)t)).
Differentiating with respect to t, we get:
v(t) = 2 + 2(cos((π/2)t) * (π/2)).
Simplifying further:
v(t) = 2 + π * cos((π/2)t).
Now, we can find the velocity at specific times t = 1 and t = 2 by substituting the values into the velocity function.
For t = 1:
v(1) = 2 + π * cos((π/2)(1)).
v(1) = 2 + π * cos(π/2).
v(1) = 2 + π * 0.
v(1) = 2.
For t = 2:
v(2) = 2 + π * cos((π/2)(2)).
v(2) = 2 + π * cos(π).
v(2) = 2 + π * (-1).
v(2) = 2 - π.
So, the velocity of the particle at t = 1 is 2 m/s, and at t = 2 is -π m/s (or approximately -3.14 m/s).
B. To determine whether the particle is speeding up or slowing down at t = 1.9 seconds, we need to consider the sign of the velocity.
Since we know that v(t) = 2 + π * cos((π/2)t), we can evaluate v(1.9) to determine if the particle is speeding up or slowing down.
v(1.9) = 2 + π * cos((π/2)(1.9)).
v(1.9) = 2 + π * cos(π(1.9/2)).
v(1.9) = 2 + π * cos(0.95π).
To evaluate the sign of cos(0.95π), we need to determine the quadrant in which 0.95π lies. Since π is half of a complete revolution (180 degrees) and 0.95π is slightly less than half, it lies in the second quadrant. In the second quadrant, cosines are negative. Therefore, cos(0.95π) < 0.
v(1.9) = 2 + (negative quantity).
v(1.9) < 2.
The velocity at t = 1.9 is less than 2 m/s. Since the velocity is decreasing, the particle is slowing down at t = 1.9 seconds.
C. To find the time t on the interval [0,3] at which the particle is farthest to the right, we need to find the maximum value of the position function s(t) on this interval.
The particle is farthest to the right when its position s(t) is at its maximum value.
s(t) = 2t + 2sin((π/2)t).
To find the maximum value of s(t), we can find the critical points of the function. We can do this by finding the values of t where the velocity, v(t), is equal to zero. At these points, the slope of the position function changes from positive to negative.
v(t) = 2 + π * cos((π/2)t).
Setting v(t) = 0 and solving for t:
2 + π * cos((π/2)t) = 0.
π * cos((π/2)t) = -2.
cos((π/2)t) = -2/π.
However, in the interval [0,3], there is no value of t that satisfies cos((π/2)t) = -2/π. Therefore, there are no critical points in this interval.
Since there are no critical points, the maximum value of s(t) on the interval [0,3] occurs at one of its endpoints.
To find which endpoint results in the maximum value, we can compare the values of s(0) and s(3).
s(0) = 2(0) + 2sin((π/2)(0)).
s(0) = 0.
s(3) = 2(3) + 2sin((π/2)(3)).
s(3) = 6 + 2sin((3/2)π).
s(3) = 6 + 2sin(3π/2).
Since sin(3π/2) = -1, we have:
s(3) = 6 + 2(-1).
s(3) = 6 - 2.
s(3) = 4.
Comparing the values, s(3) = 4 is greater than s(0) = 0.
Therefore, the particle is farthest to the right at t = 3 seconds, and its position at that time is 4 meters.
v(t) = ds/dt = 2 + πcos(π/2 t)
so now answer A with t = 1,2
B. a(t) = dv/dt = -π^2/2 sin(π/2 t)
if a(1.9) < 0 then the velocity is decreasing
C. s(t) has a maximum when v(t) = 0 and a(t) < 0
but you already know that sin(x) has a max at x = π/2 -- that is, when t=1