A particle moves along the x-axis so that at any time t>=0 its velocity is v(t)=9-t^2. If at t=1,is x=2, what is the position of the particle when it is furthest right on the x-axis .
x(t) = 9t - 1/3 t^3 + c
since x(1)=2,
2 = 9 - 1/3 + c
c = -20/3
x(t) = 9t - 1/3 t^3 - 20/3
when v=0, the particle has changed direction. That is when t=3.
So, now just evaluate x(3).
To find the position of the particle when it is furthest right on the x-axis, we need to find the maximum value of the function representing its position.
The velocity of the particle at any time t is given by v(t) = 9 - t^2.
The position of the particle at any time t can be found by integrating the velocity function. Since the velocity is the derivative of the position, we integrate v(t) with respect to t to find x(t).
∫ v(t) dt = ∫ (9 - t^2) dt = 9t - (t^3/3) + C
Now we need to find the constant C by using the initial position. We are given that when t = 1, x = 2. Plugging in these values, we get:
2 = 9(1) - (1^3/3) + C
2 = 9 - 1/3 + C
2 = 8 2/3 + C
C = -6 2/3
So the position function is:
x(t) = 9t - (t^3/3) - (6 2/3)
To find the maximum position, we need to find the critical points of the position function. These occur when the velocity is equal to zero.
v(t) = 9 - t^2
9 - t^2 = 0
t^2 = 9
t = ±3
Since the particle is moving along the x-axis, we are only interested in the positive values of t. Therefore, the critical point is t = 3.
Now we need to evaluate x(t) at this critical point:
x(3) = 9(3) - (3^3/3) - (6 2/3)
x(3) = 27 - 9 - 6 2/3
x(3) = 27 - 9 - 20/3
x(3) = 27 - 27/3 - 20/3
x(3) = 54/3 - 47/3
x(3) = 7/3
Therefore, when the particle is furthest right on the x-axis, its position is x = 7/3.