1. A particle is moving on the x-axis (or any number line) Its position x(t), or distance from the origin, at the time t is given by x(t)=4t^3-16t^2+15t. t is greater than or equal to 0
a.) Where is the particle when it is at rest?
b.) During what time interval is the particle moving to the LEFT?
c.) When is the acceleration lest?
d.) What is the TOTAL distance traveled by the particle from t=0 to t=3 ?
a) at rest means dx/dt = 0
12 t^2 -32 t + 15 = 0
t = 2.05 or .607
find x for those two values of t
b) where is dx/dt negative
do a sketch of dx(t)/dt =12t^2-32t+15
you know it is a parabola facing up (holds water) because of the positive coef of x^2
You know it is 0 at t = 2.05 and .607
so it is negative between those two times.
c) a = d^2x/dt^2 = 24 t -32
that seems to be least for t = 0 unless it means |a| which is 0 at t = 32/24
d) integral from 0 to 3 of original function is
value of t^4 - (16/3)t^3 +(15/2)t^2
when t = 3
(since it is 0 at t = 0)
To answer these questions, we will analyze the given position function step by step and use calculus concepts.
a.) To find when the particle is at rest, we need to determine when its velocity is zero. Velocity is the derivative of position with respect to time. So, we'll differentiate the given position function x(t) to get the velocity function v(t):
v(t) = x'(t) = d/dt (4t^3 - 16t^2 + 15t)
Differentiating term by term using the power rule, we get:
v(t) = (3 * 4t^2) - (2 * 16t) + 15
v(t) = 12t^2 - 32t + 15
Now, we can set v(t) equal to zero and solve for t:
12t^2 - 32t + 15 = 0
Using either the quadratic formula or factoring, we find that t = 1/2 or t = 5/6. These are the times when the particle is at rest.
b.) To determine when the particle is moving to the left, we need to analyze its velocity. If the velocity is negative, then the particle is moving to the left. We can check the sign of the velocity function v(t) we obtained earlier.
From the equation v(t) = 12t^2 - 32t + 15, we can see that it is a parabolic function opening upward. To find the time intervals when the particle is moving to the left, we need to determine when v(t) is negative. We can do this by finding the roots of the equation 12t^2 - 32t + 15 = 0 and plotting the parabola.
Using the quadratic formula or factoring, we find two roots, t ≈ 0.958 and t ≈ 1.417. So, the particle is moving to the left when t is in the interval (0.958, 1.417).
c.) To determine when the acceleration is least, we need to find the time(s) when the acceleration is zero. Acceleration is the derivative of velocity with respect to time. So, we differentiate the velocity function v(t) obtained earlier.
a(t) = v'(t) = d/dt (12t^2 - 32t + 15)
Differentiating term by term, we get:
a(t) = 24t - 32
Setting a(t) = 0 and solving for t gives:
24t - 32 = 0
t = 4/3
So, the acceleration is least at t = 4/3.
d.) To find the total distance traveled by the particle from t = 0 to t = 3, we need to consider both the positive and negative displacements.
To find the displacement from t = 0 to t = 3, we can find the position at these two times and calculate the difference:
x(3) - x(0)
Using the position function x(t) = 4t^3 - 16t^2 + 15t, we have:
x(3) = 4(3^3) - 16(3^2) + 15(3)
x(0) = 4(0^3) - 16(0^2) + 15(0)
Evaluating these expressions, we find:
x(3) = 81
x(0) = 0
Therefore, the total distance traveled by the particle from t = 0 to t = 3 is |81 - 0| = 81 units.