Are my answers right? If they are wrong you don't have to tell me how to do it its okay =)
A particle starts at time t=0 and moves along the x-axis so that its position at any time t(greater or equal to) 0 is given by x(t)=(t-1)^3 (2t-3)
1.find the velocity of the particle at any time t(greater or equal to) 0
2.for what values of t is the velocity of the particle less than zero?
3.find the value of t when the particle is moving and the acceleration is zero
My Answers:
1.(6t-9)(t-1)^2 + 2(t-1)^3
2.t(less than or equal to) 1
3.t=1
P1=(-3,2);P2=(-6,-2)
To check if your answers are correct, let's go through each question and see if your answers match the correct solutions.
1. The velocity of a particle is the derivative of its position with respect to time. In this case, you need to take the derivative of the position function x(t)= (t-1)^3 * (2t-3) with respect to t. So, the velocity function will be:
v(t) = d/dt (x(t)) = d/dt [(t-1)^3 * (2t-3)]
To find the derivative, you can apply the product rule and the chain rule. After calculating, you should get:
v(t) = (6t-9)(t-1)^2 + 2(t-1)^3
Your answer for the velocity function is correct!
2. To find the values of t for which the velocity is less than zero, you need to solve the inequality v(t) < 0. From your velocity function, v(t) = (6t-9)(t-1)^2 + 2(t-1)^3. To solve this inequality, set v(t) < 0 and find the values of t that satisfy it.
It seems that you entered t ā¤ 1, but that is incorrect. The correct solution is t < 1. So, the velocity of the particle is less than zero for t < 1.
3. To find the value of t when the particle is moving and the acceleration is zero, you need to find the values of t at which the derivative of the velocity function is zero. In other words, solve the equation for a(t) = d/dt(v(t)) = 0.
To find the derivative of the velocity function, you can apply the chain rule and simplify the expression:
a(t) = d/dt[(6t-9)(t-1)^2 + 2(t-1)^3]
= 12(t-1) + (6t-9)(2t-2)
Now you can set a(t) = 0 and solve for t:
12(t-1) + (6t-9)(2t-2) = 0
Simplifying this equation will give you the value of t when the particle is moving and the acceleration is zero.
So, to summarize:
1. Your answer for the velocity function is correct: v(t) = (6t-9)(t-1)^2 + 2(t-1)^3.
2. The correct solution for the velocity being less than zero is t < 1.
3. You need to solve the equation 12(t-1) + (6t-9)(2t-2) = 0 to find the value of t when the particle is moving and the acceleration is zero.
So, your answers for questions 1 and 3 are correct. However, the answer for question 2 should be t < 1, not t ā¤ 1.