Introductory Level Calculus Question

[Wormaldson]

I know this is the Precalculus Mathematics forum, but this is the kind of intro-level stuff that is taught in general high school mathematics classes before high school calculus, so I thought it would be better off here.

Homework Statement

An object is moving in a straight line.
P is a point on the line.
Its acceleration, a ms^–2 at a time t seconds after it leaves P, is given by
a(t) = 2 – 6t
When t = 0, the object is at point P and has a velocity of 4 ms^–1.
How far is the object from P when t = 3?

Homework Equations

a(t) = 2 – 6t

The Attempt at a Solution

Honestly, I'm not really too sure to begin with this one. It's obvious enough that P is at the y-intercept, +2, and that the gradient of a(t) = 2 – 6t is 6, but I don't know where to go from here.

Any help would be much appreciated.

 

[Mentallic]

Is it really taught as intro before calculus? Because this problem needs calculus to be solved.

The derivative of displacement is velocity and the derivative of velocity is acceleration, so going backwards, you would have to take the integral of the acceleration to get velocity and so on.

 

[Wormaldson]

Is it really taught as intro before calculus? Because this problem needs calculus to be solved.

The derivative of displacement is velocity and the derivative of velocity is acceleration, so going backwards, you would have to take the integral of the acceleration to get velocity and so on.

What I meant was, we get taught basic stuff like power rule and constant rule differentiation, some basic stuff about integration, and have a paper in our end of year exam that requires us to apply those techniques to solve problems.

Thanks for the help.

 

[Mentallic]

Oh I see, they just give you a little taste of calculus so as not to scare you off

 

[Wormaldson]

Alright, so I've got that the integral of a(t) = 2 – 6t is 2t - 3t² by reversing the power rule.

So I guess the velocity beyond point P is given by 2t - 3t² + 4 (since the velocity at P is already 4). Now I can't think of what to do next...

 

[fss]

You need to integrate the velocity function to get the position function, which will allow you to calculate x(3).

 

[Mentallic]

It asks you how far the object is from P at t=3. In other words, what is the displacement at t=3?

 

[Wormaldson]

Okay, so integrating 2t - 3t² + 4 gave me t² - t³ + 4t, which equals -6 when t = 3. I'm not entirely sure that's my final answer though; I'm not sure that I did the right thing with the +4 velocity the object has at point P.

 

[Mentallic]

When you take the derivative of some function, any constant you have is wiped clean.

For y=x², dy/dx=2x
y=x²-2, dy/dx=2x
y=x²+c, dy/dx=2x

The derivative is always the same, so when we integrate 2x we end up with x²+c, and we don't know this constant.

When we integrated the acceleration formula, we ended up with v=2t-3t²+c
and we were given information that at t=0, we had v=4 so substituting this into the equation we end up finding c=4, so then our velocity equation is v=2t-3t²+4

integrating again, we have s=t²-t³+4t+k (I just denoted the constant something else to avoid mixing it up with the constant c before)
Now if we look carefully, we will notice that we were given the information that at t=0 we start at s=0, so substituting this in we get c=0, thus our displacement formula is s=t²-t³+4t

So yes, you end up with s=-6 at t=3, which means after 3 seconds the particle would have turned around and ended up 6 units on the other end of the starting point. This can be observed intuitively by noticing that the acceleration formula a(t) = 2 – 6t, becomes negative for t>2/3 and quickly grows quite large negatively.

 

[Wormaldson]

When you take the derivative of some function, any constant you have is wiped clean.

For y=x², dy/dx=2x
y=x²-2, dy/dx=2x
y=x²+c, dy/dx=2x

The derivative is always the same, so when we integrate 2x we end up with x²+c, and we don't know this constant.

When we integrated the acceleration formula, we ended up with v=2t-3t²+c
and we were given information that at t=0, we had v=4 so substituting this into the equation we end up finding c=4, so then our velocity equation is v=2t-3t²+4

integrating again, we have s=t²-t³+4t+k (I just denoted the constant something else to avoid mixing it up with the constant c before)
Now if we look carefully, we will notice that we were given the information that at t=0 we start at s=0, so substituting this in we get c=0, thus our displacement formula is s=t²-t³+4t

So yes, you end up with s=-6 at t=3, which means after 3 seconds the particle would have turned around and ended up 6 units on the other end of the starting point. This can be observed intuitively by noticing that the acceleration formula a(t) = 2 – 6t, becomes negative for t>2/3 and quickly grows quite large negatively.

Thank you very much for taking the time to explain how to do this question to me, especially for the long explanation in the quoted post. I understand the mathematics behind this question a lot better now, which will definitely help me for my upcoming exams.

 

[Mentallic]

You're welcome good luck in your exams!