position function is x(t)=(t-1)^3 (2t-3)
how do you find the velocity of the particle at nay time when t is greater or equal to 0?
i got the derivative equation (6t-9)(t-1)^2 + 2(t-1)^3
but that is for all t.
Your formula is correct. All you have to do is apply if for positive t only.
To find the velocity of the particle at any time when t is greater or equal to 0, we need to take the derivative of the position function with respect to time.
Given the position function x(t) = (t-1)^3 * (2t-3), we can find the velocity function by taking the derivative of x(t) with respect to t.
Let's go step by step:
Step 1: Expand the expression (t-1)^3 * (2t-3):
x(t) = (t-1)^3 * (2t-3)
= (t-1)(t-1)(t-1) * (2t-3)
= (2t^2 - 6t + 3)(t-1)
Step 2: Apply the product rule to take the derivative:
To differentiate the expression (2t^2 - 6t + 3)(t-1), we can use the product rule. The product rule states that for functions u(t) and v(t), the derivative of their product is given by:
d/dt(u(t) * v(t)) = u'(t) * v(t) + u(t) * v'(t)
In our case, we have u(t) = 2t^2 - 6t + 3 and v(t) = t-1.
Applying the product rule, we get:
x'(t) = (2t^2 - 6t + 3)'(t-1) + (2t^2 - 6t + 3)(t-1)'
Step 3: Differentiate each term:
To find the derivative of each term, we need to apply the power rule and chain rule.
First term: (2t^2 - 6t + 3)' = 4t - 6
Second term: (t-1)' = 1
Step 4: Substitute the derivatives back into the equation:
x'(t) = (4t - 6)(t-1) + (2t^2 - 6t + 3)(1)
= (4t^2 - 10t + 6) + (2t^2 - 6t + 3)
= 6t^2 - 16t + 9
So, the velocity function of the particle is v(t) = 6t^2 - 16t + 9.
Therefore, at any time t when t is greater or equal to 0, the velocity of the particle is given by v(t) = 6t^2 - 16t + 9.