Rectilinear Motion:
*Need help with these three!*
Directions: The position function of a particle moving on a coordinate line is given by the following eq'ns, where s is in feet and t is in sec. Describe the motion of the particle for any time. Make a DIAGRAM of the motion with respect to its position at time t.
1.) s(t) =t^3 -6t^2
3.) s(t) = 1+6t -t^2
5.) s(t)=t^3 -9t^2 +24t
#1 is t^2(t-6)
ds/dt = 3t^2-12t = 3t(t-4)
Since s(t) it has a double root at t=0, it starts by moving down, then up again at t=4.
#2 is just a parabola, with s(0) = 1
since ds/dt > 0 at t=0, s increases until t=3, then decreases from there
#3 ds/dt = 3t^2-18t+24 = 3(t-2)(t-4)
so, s(0) = 0, and s increases while 0<t<2, decreases where 2<t<4, then increases again from there.
Go to any good online graphing site and you can see the graphs.
To describe the motion of the particle and make a diagram of its position at time t, we need to understand the position function and analyze its characteristics.
1.) s(t) = t^3 - 6t^2
First, let's determine the velocity function by taking the derivative of the position function:
v(t) = s'(t) = 3t^2 - 12t
Next, we find the acceleration function by taking the derivative of the velocity function:
a(t) = v'(t) = 6t - 12
Now, let's analyze the motion:
- To find the critical points, we set the velocity function equal to zero and solve for t:
3t^2 - 12t = 0
t(3t - 12) = 0
t = 0 or t = 4
- The particle experiences a change in direction at these critical points. At t = 0, the particle changes from moving in one direction to moving in the opposite direction, and at t = 4, the particle changes direction again.
- To find the points of inflection, we set the acceleration function equal to zero and solve for t:
6t - 12 = 0
t = 2
- The point of inflection occurs at t = 2.
Now, let's create a diagram to visualize the motion:
- On a graph with t on the x-axis and s(t) on the y-axis, plot the following key points:
- (0, 0) - starting point
- (2, s(2)) - point of inflection
- (4, s(4)) - point of direction change
- Connect these points with smooth curves, considering the shape of the position function and the behavior at critical points and the point of inflection.
Repeat the same process for the other position functions:
3.) s(t) = 1 + 6t - t^2
Find the velocity function:
v(t) = s'(t) = 6 - 2t
Find the acceleration function:
a(t) = v'(t) = -2
Analyze the motion:
- Since the acceleration is a constant -2, the particle undergoes uniform deceleration.
5.) s(t) = t^3 - 9t^2 + 24t
Find the velocity function:
v(t) = s'(t) = 3t^2 - 18t + 24
Find the acceleration function:
a(t) = v'(t) = 6t - 18
Analyze the motion:
- To find the critical points, we set the velocity function equal to zero and solve for t:
3t^2 - 18t + 24 = 0
t^2 - 6t + 8 = 0
(t - 2)(t - 4) = 0
t = 2 or t = 4
- The particle changes direction at these critical points.
- To find the points of inflection, we set the acceleration function equal to zero and solve for t:
6t - 18 = 0
t = 3
- The point of inflection occurs at t = 3.
Create a diagram by plotting the key points and connecting them with smooth curves, considering the behavior of the position function, critical points, and the point of inflection.
Remember, to get the most accurate representation of the motion, it is always helpful to plot more points and consider the behavior of the functions within the given time intervals.