The position of a particle in meters is given by x =3.6t2 - 2.4t3 , where t is time in seconds.
A) Find the two times when the velocity is zero.
B) Calculate the position for each of those times.
V = dx/dt = 7.2 t - 7.2 t^2
= 7.2t*(1-t)
A) V = 0 when t = 0 and t = 1 s
B) Susbtitute those times in the x(t) equation.
To find the times when the velocity is zero, we need to find the value(s) of t that make the derivative of the position equation (x) equal to zero.
Let's begin by finding the derivative of the position equation with respect to time (t):
x = 3.6t^2 - 2.4t^3
To find the derivative, we can differentiate each term separately:
dx/dt = d(3.6t^2)/dt - d(2.4t^3)/dt
Using the power rule of differentiation, where d(x^n)/dt = nx^(n-1), we can find the derivatives:
dx/dt = 7.2t - 7.2t^2
Now, we need to set dx/dt equal to zero and solve for t:
0 = 7.2t - 7.2t^2
Now, let's factor out t:
0 = 7.2t(1 - t)
From here, we have two possibilities:
1) The first factor is zero: 7.2t = 0. This gives us t = 0.
2) The second factor is zero: 1 - t = 0. Solving for t, we get t = 1.
So, we have two times when the velocity is zero: t = 0 and t = 1.
Now, let's calculate the position for each of these times:
A) For t = 0:
x = 3.6(0)^2 - 2.4(0)^3
x = 0
The position at t = 0 is x = 0 meters.
B) For t = 1:
x = 3.6(1)^2 - 2.4(1)^3
x = 3.6 - 2.4
x = 1.2
The position at t = 1 is x = 1.2 meters.