(a) If the position of a particle is given by x=20t-5t^{3}, where x is in meters and t is in seconds, when, if ever, is the particle’s velocity zero? (b) When is its acceleration a zero? (c) For what time range (positive or negative) is a negative? (d) Positive? (e) Graph x(t), v(t), and a(t).

#### Solution:

### The derivative of a position function will give us the velocity and the second derivative of a position function, or the first derivative of a velocity function will give us the acceleration function.

### Thus, let us first take the derivative of the position function.

### v(t)=\frac{dx(t)}{dt}=-15t^{2}+20

### Let us now take the derivative of the velocity function we just obtained.

### a(t)=\frac{dv(t)}{dt}=-30t

### From these equations, we will now answer the questions.

### a) To figure out when the velocity of the particle is zero, we simply find t that gives us 0 as it’s answer. Therefore:

### 0=-15t^{2}+20

### t=\sqrt{\frac{20}{15}}

### t=1.2 s

### b) As before, to find when the acceleration is zero, we need to find the value of t that gives us 0 as it’s answer in our function. Therefore:

### 0=-30t

### t=0

### c) We can see that because our acceleration, a=-30t, for any value of t>0, we would get a negative value.

### d) We can also see that for any value of t<0, we would get a positive acceleration value.

### e)

###### This is question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 20.