assume nonconstant acceleration

[logistic_guy]

Suppose that a car can accelerate from 30 mph to 50 mph in 4 seconds. Assuming nonconstant acceleration, find the acceleration (in miles per second squared) of the car and find the distance traveled by the car during the 4 seconds.

this not book question. i just want to know the difference of constant and nonconstant acceleration
i'm certain the 3 formulas can't be used this time
$\displaystyle v = v_0 + at$
$\displaystyle x = x_0 + v_0t + 0.5at^2$
$\displaystyle v^2 = v_0^2 + 2ax - 2ax_0$

how to solve this question? can't be solved with this assumption?

 

[blamocur]

To answer this question you need to know more about acceleration. I.e., how exactly it varies with time?

 

[logistic_guy]

give me example of acceleration varies with time

 

[Steven G]

give me example of acceleration varies with time

a(t) = 4t^2
a(t) = -3t^2 + 9
a(t) = 9.8t^2
....

 

[topsquark]

Suppose that a car can accelerate from 30 mph to 50 mph in 4 seconds. Assuming nonconstant acceleration, find the acceleration (in miles per second squared) of the car and find the distance traveled by the car during the 4 seconds.

this not book question. i just want to know the difference of constant and nonconstant acceleration
i'm certain the 3 formulas can't be used this time
$\displaystyle v = v_0 + at$
$\displaystyle x = x_0 + v_0t + 0.5at^2$
$\displaystyle v^2 = v_0^2 + 2ax - 2ax_0$

how to solve this question? can't be solved with this assumption?

The question is too vague. You have to specify some form for the acceleration (are you thinking linear in time?) to do the problem.

-Dan

 

[logistic_guy]

post 4 how this help me to find the acceleration in post 1

post 5, look at post 4 he give us the form for the acceleration

 

[The Highlander]

Suppose that a car can accelerate from 30 mph to 50 mph in 4 seconds. Assuming nonconstant acceleration, find the acceleration (in miles per second squared) of the car and find the distance traveled by the car during the 4 seconds.

this not book question. i just want to know the difference of constant and nonconstant acceleration
i'm certain the 3 formulas can't be used this time
$\displaystyle v = v_0 + at$
$\displaystyle x = x_0 + v_0t + 0.5at^2$
$\displaystyle v^2 = v_0^2 + 2ax - 2ax_0$

how to solve this question? can't be solved with this assumption?

Understanding the complexities of the questions you pose requires an understanding of Physics. Let me try explaining (what I think is) your query by referring to the graph I have created below.

The graph is a velocity (or speed?) versus time graph (for the purposes of this explanation I will use the terms velocity & speed interchangeably even though in Physics they are regarded as two very different things!).

The x-axis shows time in seconds and the y-axis speed/velocity in miles/hour. (Not a unit we would often use in Physics questions beyond a very early level).

Your driver starts off at 30mph and at the 4 second mark s/he accelerates, over a 4 second period, to 50 mph.
This is the situation described in your OP.

The green line shows a nice smooth (constant) acceleration where the speed of the car is increasing by 5mph every second; so it reaches 35mph after 1s, 40mph after 2s, 45mph after 3s and stops accelerating after 4s having now reached 50mph and continues at that speed.

Note that the lines on the graph are showing the vehicles speed (velocity) and so when the line is flat (horizontal) that indicates a steady (constant) speed but the rising line (between 4 & 8 seconds) is showing that the speed is changing, ie: that there is an acceleration, but the graph is still plotting speed over that time period (it is not plotting acceleration).

However, the driver might just as easily have 'floored' the gas pedal at the 4s mark to get the maximum acceleration out of the car then eased up on the pedal after 2s (having reached c.40mph) such that the speed was still increasing (but more slowly) until 50mph was reached. That kind of behaviour might be represented by the red line I have added to the graph.

So the red line would represent a non-constant (I would call it variable) acceleration.

​

You appear to be asking whether the formulas you quote might be used in this situation and the answer is: No!

For a start, the formulas you list are (as far as I am concerned) not correct! I already referred you to a post where I explained their (proper form &) use previously Here.

If you look at that post you will see that the formula used in Physics is: $\displaystyle s=ut+\frac{1}{2}at^2$

the equation you quote ($\displaystyle x = x_0 + v_0t + 0.5at^2$) is inconsistent because an x₀ and a v₀ term would imply a displacement and velocity at a concurrent time; the x₀ term simply shouldn't be there at all as the equation is intended to calculate displacement (distance to you) over a specific time interval and any preceding displacement should be added afterwards (if necessary).

The correct form of this equation (in Physics) is what I have stated above; using s for displacement, u for initial velocity, a for (constant) acceleration and t for time to calculate the displacement (distance travelled to you) over a specific time interval.

If you have been given the equations you quote in, perhaps, a Maths class then they will be what you are expected to use but how to use them (in that environment) should have been clearly explained to you first!

Now, why can't they be used under variable acceleration?

Well, look at the graph again. In order to calculate a distance when you know speed & time you multiply the speed by the time.

$$\sf Speed =\frac{Distance}{Time}\implies Distance=Speed\times Time$$
So to find a distance travelled from a speed-time graph you find the AREA under the graph during the time you are interested in. That is effectively multiplying the speed over a specific time interval by the time interval itself, thus giving the distance travelled during that time interval.

Now that's fine when the speed is constant; eg: in the first 4s the car travels at a constant 30mph (1/120 mi/sec) so 1/120 x 4 = 0.03... miles (58.6... yards or 176 feet).

But when the velocity (speed) is changing then things become more complicated and you can no longer just multiply speed by time (because the speed is now changing over time).

Now (for the green line) you need to find the area of the rectangle shown by dotted lines in the graph plus the area of the triangle above it.

When the acceleration is constant (as it is in the green line) then the equation $\displaystyle s=ut+\frac{1}{2}at^2$ does this for you.

The $\displaystyle ut$ part calculates the area of the rectangle and the $\displaystyle \frac{1}{2}at^2$ part calculates the area of the triangle.

You may remember that the very first response you got in your original thread was from @Dr.Peterson where he said: "This is really a calculus problem (though it could also be assigned in a physics course where you are just given a formula)." He is absolutely correct.

When the acceleration is constant (ie: velocity is increasing at a constant rate) then the the v-t graph will have a straight line with a constant slope and when that is integrated (ie: using Calculus to find the area under the graph) it produces the expression: $\displaystyle \frac{1}{2}at^2$ because the acceleration (a) is the derivative of the speed with respect to time and integration is the anti-derivative so it takes you back to the (accumulative) velocity over that time interval.

The area above the rectangle is now no longer a nice neat triangle but is now enclosed by a wavy line!

In order to calculate the distance travelled over the 4 seconds when the speed varied in accordance with the red line, you would need to know the equation of the red line and be able to integrate that over the time period concerned to get the area (now) above the rectangle.

 

[blamocur]

give me example of acceleration varies with time

$a(t) = \cos (t)$

 

[khansaheb]

$a(t) = \cos (t)$

A swinging pendulum will have acceleration of that form.

 

[Steven G]

post 4 how this help me to find the acceleration in post 1

You asked give me example of acceleration varies with time. I did just that because I thought that you wanted examples so that you could think about what all this means.

 

[logistic_guy]

post 7 thank you for the explanation. i understand now difference of consant and nonconstant acceleration. question in post 1 isn't good to find nonconstant acceleration because graph is describing velocity and i need to know nonconstant velocity to find the nonconstant acceleration.

if i need distance, i find area under curve of velocity graph and if i need acceleration i find slope of velocity graph. this is my understand and i think it make sense since $\displaystyle v = \frac{dx}{dt}$ and $\displaystyle a = \frac{dv}{dt}$

i'm sorry to say this, but i have never seen your formulas in my physics class. in physics class we use
$\displaystyle v = v_0 + at$
$\displaystyle x = x_0 + v_0t + 0.5at^2$
$\displaystyle v^2 = v_0^2 + 2ax - 2ax_0$
physics book use this
$\displaystyle v = v_0 + at$
$\displaystyle x = x_0 + v_0t + \frac{1}{2}at^2$
$\displaystyle v^2 = v_0^2 + 2a(x - x_0)$
$\displaystyle \bar{v} = \frac{v + v_0}{2}$

i'm sure this is physics problem, not calculus. i'll show a picture next time of the formulas notation and question

i'm also confused at post 2, he say we need to know how exactly acceleration varies with time to answer the question, why he didn't say we need to know how exactly velocity varies with time time instead. may got confused?!

i'm an artist. i draw people anime nature any thing. i also enter a house and i can design a similar small version of it. i'm currently studying mechanical engineer major and chemical engineering minor. i'm very bad in beginning math courses like algebra. this make struggling in main courses. this is why i memorizing to solve question without understand any math

i just want to understand three things. basic idea of algebra, basic idea of physics, and basic idea of calculus

 

[blamocur]

post 7 thank you for the explanation.

A better and simpler way to address particular person would be by their user name instead of the post number, which has to be looked up. For example, post 7 is also known as @The Highlander.

 

[logistic_guy]

Highlander, i can't find question in physics book, but i show picture of formuals notation



please use the forum's feature for quoting content. please what is this mean?

 

[topsquark]

Highlander, i can't find question in physics book, but i show picture of formuals notation

The equations you are posting are only for constant acceleration, which is the case that you had in your original problem. If acceleration is not constant then we need to specify the form of that acceleration and rederive everything.

-Dan

 

[khansaheb]

then we need to specify the form of that acceleration and rederive everything.

@logistic_guy

Do you know how the original 3 equations were derived?​

​

If you don't - please dust-up your old physics book and get yourself acquainted with the process.​

​

That will give you the process to derive equivalent equations for "non-constant acceleration".

 

[The Highlander]

The equations you are posting are only for constant acceleration, which is the case that you had in your original problem. If acceleration is not constant then we need to specify the form of that acceleration and rederive everything.

-Dan

I provided a detailed explanation of this at Post #7 and, I'm happy to say , s/he now seems to understand.

 

[The Highlander]

Highlander, i can't find question in physics book, but i show picture of formuals notation

Thank you for posting your picture and I can see where you are getting your formulas from now but, as I said earlier: "If you have been given the equations you quote...then they will be what you are expected to use but how to use them (in that environment) should have been clearly explained to you first."

Just the page you have provided repeatedly warns that a must be constant to use the formulae...

​

please use the forum's feature for quoting content. please what is this mean?

I have no idea what that means unless it is a suggestion that you use the "Reply" link at the bottom RHS if you want to include something from a previous post (like I have done above).

Can I just say: It's very important that you give a lot of thought to what you want to know and how you phrase your posts. I understand that English is not your first language but we can only respond to what you actually write!

Regards,
TH

 

[logistic_guy]

Thank you for posting your picture and I can see where you are getting your formulas from now but, as I said earlier: "If you have been given the equations you quote...then they will be what you are expected to use but how to use them (in that environment) should have been clearly explained to you first."

Just the page you have provided repeatedly warns that a must be constant to use the formulae...

View attachment 37906​

so it is difficult to understand what made you ask if they could be used when a was "nonconstant" but I trust you now understand why a must be constant to use them.


I have no idea what that means unless it is a suggestion that you use the "Reply" link at the bottom RHS if you want to include something from a previous post (like I have done above).

Can I just say: It's very important that you give a lot of thought to what you want to know and how you phrase your posts. I understand that English is not your first language but we can only respond to what you actually write!

If you ask the wrong question or ask the right question wrongly, then you won't get the information you want and then threads grow far too long as you ask more questions trying to get at what you want. Helpers then become exasperated too when their responses just produce more questions.

Regards,
TH

i send formulas, not because i don't understand. you say the formula $\displaystyle x = x_0 + v_0t + 0.5at^2$ is inconsistent. this is what we're using in class

thank you The Highlander, i solve many questions, but first time to understand the graph of physics. i think i understand this question and previous. i understand what i wanted. thank you The Highlander

thank you also about the reply link. i don't see it before