The position of a particle moving along the x-axis as a function of time,t, is given by x(t)=(1/6)t^3-t^2+3t-1 for t≥0. The particle's velocity becomes three times its initial velocity when t=?
I know v(t)=x'(t)=(1/2)t^2-2t+3=9, but I do not understand where does the 9 come from.
Some help would be appreciated, thanks.
read the question carefully.
v(0) = 3
you want v to be three times its initial value. 3*3 = 9
To find the value of t when the particle's velocity becomes three times its initial velocity, we need to set the velocity equation equal to three times the initial velocity and solve for t.
You correctly found the velocity equation: v(t) = (1/2)t^2 - 2t + 3.
Now, to determine the initial velocity, we need to find the derivative of the position equation x(t). Taking the derivative of x(t), we get x'(t) = (1/2)3t^2 - 2(2t) + 3 = (3/2)t^2 - 4t + 3.
Since the initial velocity is the velocity when t = 0, we can substitute t = 0 into the velocity equation: v(0) = (3/2)(0)^2 - 4(0) + 3 = 3.
So the initial velocity is 3, and now we can set up the equation:
(1/2)t^2 - 2t + 3 = 3(3).
Simplifying this equation, we get:
(1/2)t^2 - 2t + 3 = 9.
Subtracting 9 from both sides, we have:
(1/2)t^2 - 2t - 6 = 0.
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a).
In this case, a = 1/2, b = -2, and c = -6.
Plugging in these values, we get:
t = (-(-2) ± √((-2)^2 - 4(1/2)(-6))) / (2(1/2))
= (2 ± √(4 + 12)) / 1
= (2 ± √16) / 1
= (2 ± 4) / 1.
This gives us two possible values for t:
t = 2 + 4 = 6, and
t = 2 - 4 = -2.
However, since we are given that t is greater than or equal to 0, the only valid solution is t = 6.
So the particle's velocity becomes three times its initial velocity when t = 6.