Introductory Level Calculus Question

In summary, the conversation discusses a problem involving an object moving in a straight line, with its acceleration given by the formula a(t) = 2 – 6t at time t seconds. The object has a velocity of 4 ms–1 at point P when t=0. The conversation explores how to find the object's distance from P when t=3 using calculus techniques. It is mentioned that these techniques are typically taught at an introductory level in high school mathematics classes before calculus. The conversation ends with a clear explanation of how to solve the problem, including finding the displacement formula s=t2-t3+4t and determining that the object would have turned around and ended up 6 units on the other end of the starting point
  • #1
Wormaldson
21
0
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. :confused:

Any help would be much appreciated.
 
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  • #2
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.
 
  • #3
Mentallic said:
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.
 
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  • #4
Oh I see, they just give you a little taste of calculus so as not to scare you off :wink:
 
  • #5
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...
 
  • #6
You need to integrate the velocity function to get the position function, which will allow you to calculate x(3).
 
  • #7
It asks you how far the object is from P at t=3. In other words, what is the displacement at t=3?
 
  • #8
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.
 
  • #9
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.
 
  • #10
Mentallic said:
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.
 
  • #11
You're welcome :smile: good luck in your exams!
 

1. What is calculus?

Calculus is a branch of mathematics that deals with the study of change and motion. It involves the use of mathematical tools and techniques to solve problems related to rates of change, optimization, and approximation.

2. What is the difference between differential and integral calculus?

Differential calculus involves the study of rates of change, while integral calculus deals with the accumulation of quantities over a given interval. In simpler terms, differential calculus is concerned with finding the slope of a curve, while integral calculus focuses on finding the area under a curve.

3. What are the basic concepts of calculus?

The basic concepts of calculus include limits, derivatives, and integrals. Limits are used to describe the behavior of a function as the input values approach a certain value. Derivatives are the instantaneous rates of change of a function, while integrals are used to find the area under a curve.

4. How is calculus used in real life?

Calculus has various applications in real life, such as in engineering, physics, economics, and statistics. It is commonly used to model and solve problems related to motion, optimization, and growth. For example, it can be used to determine the maximum profit a company can make or to calculate the speed of a moving object.

5. Is calculus difficult to learn?

Calculus can be challenging to learn, but it is also a very rewarding subject. It requires a strong foundation in algebra and trigonometry, as well as critical thinking and problem-solving skills. With dedication and practice, anyone can learn and excel in calculus.

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