1-The position of a particle moving along the x axis is given by x=t^2-〖2t〗^3-6, where x is in meters and t in seconds. What is the position of the particle when it achieves its maximum speed in the positive x direction?
x = t^2 - (2t)^3 - 6
v = 2t - 16t^2
v is maximum when a=dv/dt=0
a = 2 - 32t
now find t when a=0. Since da/dt < 0, v will be a maximum.
How you find - 16t^2
To find the position of the particle when it achieves its maximum speed in the positive x direction, we need to find the maximum speed and then substitute that time into the position equation.
Step 1: Find the velocity of the particle.
The velocity of the particle, v, is given by the derivative of the position equation with respect to time, t:
v = dx/dt
Differentiating the position equation, we get:
v = d/dt (t^2 - 2t^3 - 6)
v = 2t - 6t^2
Step 2: Find the time when the velocity is maximum.
To find the maximum speed, we need to find the time when the velocity is maximum. This can be found by setting the derivative of the velocity equation to zero and solving for t:
dv/dt = 2 - 12t = 0
Solving for t, we get:
2 - 12t = 0
12t = 2
t = 2/12
t = 1/6
Step 3: Substitute the time into the position equation.
Substituting t = 1/6 into the position equation: x = t^2 - 2t^3 - 6
x = (1/6)^2 - 2(1/6)^3 - 6
x = 1/36 - 2/216 - 6
x = 1/36 - 1/108 - 6
x = (1 - 3 - 216)/216
x = -218/216
x = -1.009
Therefore, the position of the particle when it achieves its maximum speed in the positive x direction is approximately -1.009 meters.