+223 The position of a particle is given by x(t) = (t2 - 4)3 where x(t) is measured in centimeters and t is measured in seconds.
Find where the particle is moving to the left.
So I know if I take the derivative of this equation that it will represent the velocity:
6t3 - 24t = v(t)
Now I also know that when the velocity is below zero is when the particle is moving to the left. So if I plug zero into v(t) and then factor I get ----> 6t(t + 2)(t - 2)
So that means that the slope of the graph (the velocity) is zero at t = 0 t = -2 and t = 2
But without using a graphing calculator, how am I supposed to know when the slope is negative? As far as I can tell, there is nowhere after t = 0 that the slope ever goes negative again. At t = 2 for example, the slope just approaches zero and then it is positive again. It seems that the particle never moves left.
So does the particle ever move left or not? And if so, where?
+118677 The position of a particle is given by x(t) = (t2 - 4)3 where x(t) is measured in centimeters and t is measured in seconds.
\(x(t)=(t^2-4)^3\)
Find where the particle is moving to the left.
velocity is negtive
\(x(t)=(t^2-4)^3\\ x'(t)=3(t^2-4)^2*2t\\ x'(t)=6t(t^2-4)^2\\ \)
Since t cannot be negative, velocity can never be negative.
+118677 The position of a particle is given by x(t) = (t2 - 4)3 where x(t) is measured in centimeters and t is measured in seconds.
\(x(t)=(t^2-4)^3\)
Find where the particle is moving to the left.
velocity is negtive
\(x(t)=(t^2-4)^3\\ x'(t)=3(t^2-4)^2*2t\\ x'(t)=6t(t^2-4)^2\\ \)
Since t cannot be negative, velocity can never be negative.
