Please help me solve this calculus question.

Question

The motion of the particle is described by the position function :

S ( t) = t^3 – 9t^ 2 + 24^t + 5; t ≥ 0

Where t is measured in seconds and s(t) in meters

a) At what time is a particle at rest?

b) When is the object moving in a negative direction?

c) Is the particle moving away or towards the origin when t = 1 second? Justify the answer.

d) Is the particle speeding up or slowing down when t=1 second ?

e) When is its acceleration positive ?

Answer 1

We'll need the derivative of S(t) fairly extensively, so let's go ahead and calculate it now.

S'(t)=3t²-18t+24

Looking ahead, it will be useful to factor this expression into a product of binomials.

S'(t)=3t²-18t+24

S'(t)=3[t²-6t+8]

S'(t)=3[t²-2t-4t+8]

S'(t)=3[t(t-2)-4(t-2)]

S'(t)=3(t-4)(t-2)

That's what it is mathematically, but physically, what does this represent? Well, remember that the derivative is the rate of change of a function, and since the original function was position with time, this should tell you how the position is changing with time (i.e. velocity if you're familiar with physics). For instance, if you find a value of t such that S'(t) is positive, that tells you that at that moment the particle is moving to the right (because it's position is increasing with time).

So, for part a, we need to find when the particle is at rest. Well, to be at rest means to not be moving, meaning the particle can't be moving to the left or to the right. Mathematically, that means that S'(t) must be 0. So, when is that expression 0? Well, because we cleverly factored it ahead of time, we know that S'(t) is 0 when t=2 or t=4!

What about b? When is the particle moving in a negative direction? Well, that happens when S'(t) is negative, so we need to analyze for when that happens. Since we have it expressed as a product of terms, we can recognize that S'(t) is negative either when one or all of its terms are negative (if only two of them are negative, then the product will still be positive. Likewise, if all are positive, the product is again positive). Since the first term is just the number 3, which is always positive, it's impossible for all three terms to be negative. So, we're restricted to just finding when one and only one of the t-4 and t-2 is negative.From inspection, we can find that in the range (2,4), t-2 will be positive while t-4 will be negative. For [4,inf], both terms are positive, and for [0,2), both terms are negative. Hence, (2,4) is the only domain where S'(t) is negative, i.e. the only values for t so that the object moves in the negative direction.

Part c we have to be a little bit careful. At t=1 second, we recognize from part b that S'(1)>0, and therefore the object is moving to the right. But that's not enough information on its own! For example, suppose the object were currently at the point x=1; because it's to the right of the origin, and currently moving right, it would be moving AWAY from the origin. On the other hand, were it at x=-1, it would be to the left of the origin but moving right, and so would be moving TOWARDS the origin. We therefore need to calculate S(1)!

S(1)=1³-9*1²+24*1+5=1-9+24+5=21

Since this value is positive, it is to the right of the origin. Therefore, since the particle is also moving right, the particle is moving AWAY from the origin.

For the last two parts, we need the acceleration function. The acceleration is the rate of change of the velocity in time, or the derivative of velocity with respect to time. Specifically, we need dS'(t)/dt, or S''(t).

S''(t)=dS'(t)/dt=d/dt (3t²-18t+24)=6t-18

A particle is speeding up if the direction of acceleration MATCHES the direction of velocity. You might find that a little counter intuitive; shouldn't the acceleration need to be positive? But, imagine a particle moving in the negative direction; if that particle were subjected to a positive acceleration, it would be adding positive velocity to negative velocity, reducing the speed (Recall that speed is the absolute magnitude of velocity, so that making velocity more negative OR more positive increases the speed). We can quickly verify that S''(1)=6*1-18=6-18=-12 (i.e. to the left), while in our work in part c we found that S'(1)>0, to the right. Because the directions of velocity and acceleration are opposite, the object is actually slowing down.

Finally, part d, when is the acceleration positive? Well, when S''(t) is positive. When is that?

S''(t)=6t-18>0 -> 6t>18 -> t>3

So for (3,inf), the acceleration will be positive.

Edit: I've just realized that in your given expression for S(t), you list it as 24^t, rather than 24*t. This would completely change the answers, but not the actual steps. You'd still want to set the derivative equal to 0 and solve for t for part a, find the values of time that make the derivative negative for b, compare the sign of S(1) and S'(1) for c, compare the sign of S'(1) and S''(1) for d, and finally find when S''(t) is positive for e. I'm leaving what I've written, because I think that you meant 24*t, since solving these equations with 24^t would be substantially more difficult.

Answer 2

let's say there is a typo and 24^t is 24t

then s'(t)=0 at t=2 and 4 seconds

s'(t)<0 2<t<4

s'(t)>0 t=1

s''(t)>0 t>3

please compute derivative and second derivative and verify arithmetic