i need to find the speed of acceleration

logistic_guy

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this is the question

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

i think the acceleration to subtract the mph so i get 30 - 50 = -20. this is negative so i think acceleration can't be negative? i am stuck with the 4 seconds but i think it is used to find the distance. it gives me two speeds, 30 mph and 50 mph. which one will be used to find the distance?

i am new here so i apologize if this was posted in the wrong place
 
i think the acceleration to subtract the mph so i get 30 - 50 = -20. this is negative so i think acceleration can't be negative?
The change in velocity is the new minus the old. You just subtracted in the wrong direction.
Suppose that a car can accelerate from 30 mph to 50 mph in 4 seconds. Assuming a constant acceleration, find the acceleration (in miles per second squared) of the car and find the distance traveled by the car during the 4 seconds.

...

i am new here so i apologize if this was posted in the wrong place
This is really a calculus problem (though it could also be assigned in a physics course where you are just given a formula). It is definitely more than arithmetic.

Can you give us some information about your background?

(This was one of many submissions that were stuck due to a difficulty we are having with moderation.)
 
i don't understand why you say velocity. isn't the question is asking for acceleration and distance?
this isn't calclus. i study calculus different things. this physics my background is art, i am an aritist
 
"i don't understand why you say velocity"
Acceleration is defined as the change in velocity over time, so by taking vf - vi and then dividing by elapsed time, you can find the acceleration.
 
but the question don't give velocity. i read this from wikipedia Velocity is a measure of how fast something moves in a particular direction. To define it needs both magnitude and direction. If an object moves east at 9 metres per second (9 m/s), then its velocity is 9 m/s to the east.

30 mph to 50 mph are speeds, not velocity. if you say velocity, please tell me what is the direction of the car?
 
but the question don't give velocity. i read this from wikipedia Velocity is a measure of how fast something moves in a particular direction. To define it needs both magnitude and direction. If an object moves east at 9 metres per second (9 m/s), then its velocity is 9 m/s to the east.

30 mph to 50 mph are speeds, not velocity. if you say velocity, please tell me what is the direction of the car?
If a car accelerates from 30 to 50 mph, then it is presumably going in the same direction at the start and end of that time. Call that the positive direction.

You are overthinking this.

i don't understand why you say velocity. isn't the question is asking for acceleration and distance?
this isn't calclus. i study calculus different things. this physics my background is art, i am an aritist
Are you saying this question is from a physics class? Then they will have given you a formula or two related to acceleration. What are they? (Those formulas are, in part, derived from calculus.)

If not, where does the question come from?
 
acceleration to subtract the mph so i get 30 - 50 = -20.
Incorrect.
Assuming that the particle moves in a straight-line, without "turning-around", the direction of the velocity does not matter. In that case,

acceleration = (change in velocity in a given time-period)/(time-elapsed) = [(final velocity) - (initial velocity)] / (time-elapsed).............(1)

In your given case,
initial velocity = 30 mph.............final velocity = 50 mph
time elapsed = 4 sec = 4/3600 hours

acceleration = (50 - 30)/(4/3600) [mph/h] = 18 000 mph2

In physics, equation (1) is (often) known as Galileo's first equation of linear motion of particles, under constant acceleration. There are two additional equations of motion for linear motion of particles under constant acceleration. You will need to use at least one of those. Consult your physics class-notes/text-book or do a Google (or Bing) search and tell us what you found.
 
i'm not overthing. i am talking logic. yo say the car go from 30 mph to 50 mph in the same direction. but that not true because the question don't say anything about the direction. what if the car car go from 30 mph to 50 mph in a curve? this is my logic we can't tell the direction because it can be anything. we studied 3 formulas involvign acceleration but i can't use them in this question because they need velocity and the question give me speed.

acceleration = (50 - 30)/(4/3600) [mph/h] = 18 000 mph2

this is what i need but i don't understand it. why 50 - 30 not 30 - 50? is it because negative we can't do that? also i don't understand division 4/3600. can you please explain?
 
why 50 - 30 not 30 - 50?
Because your initial speed is 30 mph and final speed is 50 mph.
accelerate from 30 mph to 50 mph in 4 seconds
accelerates FROM initial speed TO final speed.

what if the car car go from 30 mph to 50 mph in a curve?
Then you'll have multiple accelerations in multiple directions - things become complicated. Analyses of "general" motion of a particle - e.g. motion of a spacecraft - requires use of calculus.
 
i'm not overthing. i am talking logic. yo say the car go from 30 mph to 50 mph in the same direction. but that not true because the question don't say anything about the direction. what if the car car go from 30 mph to 50 mph in a curve? this is my logic we can't tell the direction because it can be anything. we studied 3 formulas involvign acceleration but i can't use them in this question because they need velocity and the question give me speed.
Technically, the problem is incomplete; it assumes common knowledge, as many do. There are two main ways to deal with this:

One is to observe that the problem has insufficient information, and can't be solved, if we assume that it moves in a curve; so we choose to interpret it in a way that can be solved, as it was presumably written by someone who knows the subject.

The other is to observe that we commonly talk about a car's acceleration under straight line conditions, so we can assume that (in the absence of other indications). In addition, it is common to use the words "speed" and "velocity" interchangeably, even in a physics class, where the context seems clear. Velocity is just speed in a certain direction.

In either case, the appropriate thing to do on a test would be to write out your assumptions, so that a grader will know what problem you are actually solving. (You might even get extra credit for showing this awareness.)

The wrong thing to do is to refuse to solve the problem because it is not perfect.

One more thing: Can you show us the formulas you were given, including the definitions of the variables? There will be something worth discussing there.
 
this my main point. if the initial speed 30 mph why i don't write 30 - 50. the speed 30 come first why not writing it first? for the conversion 4/60*60 is this a rule? if time in seconds do i have always to divide over 3600?

the 3 formulas v = v0 + at, x = x0 + vot + 0.5at2, v2 = v02 + 2ax - 2ax0

v is the final velocity, v0 is the initial velocity, x is the final distance, x0 is the initial distance, t is the time, a is the acceleration

there is no tap for power. it is difficult to write the power of 2 in this site
 
this my main point. if the initial speed 30 mph why i don't write 30 - 50. the speed 30 come first why not writing it first? for the conversion 4/60*60 is this a rule? if time in seconds do i have always to divide over 3600?
This is how arithmetic works. If something increases from, say, 5 to 7, the increase is the new value minus the old value, 7 - 5 = 2. This should be familiar.

And unit conversions can be explained in various ways, but one is that since one hour is 3600 seconds, the number of seconds is 3600 times the number of hours, and the number of hours is the number of seconds divided by 3600.

there is no tap for power. it is difficult to write the power of 2 in this site
I take it that, in your language, a "tap" is a button to tap, or something like that? You can read about how to enter formulas here:


Here are your formulas written as recommended there:
the 3 formulas v = v_0 + at, x = x_0 + v_0t + 0.5at^2, v^2 = v_0^2 + 2ax - 2ax_0

v is the final velocity, v0 is the initial velocity, x is the final distance, x0 is the initial distance, t is the time, a is the acceleration
Using the "f(x)" button, you can enter them in a similar but more complicated format and they look like this:
[math]v = v_0 + at\\x = x_0 + v_0t + 0.5at^2\\v^2 = v_0^2 + 2ax - 2ax_0[/math]
But also, the triple dots at the end of the menu bar opens several new buttons, including X1 and X1, for subscripts and superscripts:

v = v0+ at​
x = x0+ v0t + 0.5at2
v2= v02 + 2ax - 2ax0

(It's a little buggy currently.)

The first of these can be solved for a to find the acceleration in your problem; the second gives you the distance.
 
you're correct about the tap. i like your explanation i am understanding but it is annoying i can't express my mathematics with correct symbols. there is few points i want to understand about this question. i have to be absent the next days to learn Latex code after that we'll continue. hold on Dr.Peterson until we meet again. thanks everyone for the help.
 
this is the question

Suppose that a car can accelerate from 30 mph to 50 mph in 4 seconds. Assuming a constant acceleration,

find the acceleration (in miles per second squared) of the car and find the distance traveled by the car during the 4 seconds.

logistic_guy,

for this first question (before converting it to the required units), the acceleration =

\(\displaystyle \ \dfrac{(final \ velocity \ in \ mph) \ minus \ (initial \ velocity \ in \ mph)}{travel \ time \ in \ sec} \)

The final velocity is 50 mph.
The initial velocity is 30 mph.
The travel time is 4 sec.

However, the units for acceleration required for the answer is miles per sec per sec.

Already discussed above in at least one other post is a conversion for
miles per sec per sec.
I don't know if that discussion is finished.

---------------------------------------------

For the second question, do any of the formulas posted in post # 13 work?
Suppose this is not the case.
Since the problem is dealing with miles per hour, it would make sense to me
to find the number of feet that the car has traveled during the four seconds.
To do that, we should convert 30 mph and 50 mph to the appropriate numbers
of feet per sec each.

We could estimate the distance using rate*time by using 10 intervals, for instance:

Midpoints second marks 3.1, 3.3, 3.5, 3.7, 3.9, 4.1, 4.3, 4.5, 4.7, 4.9
Multiply these by their respective velocities and add them up.
 
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i think the acceleration to subtract the mph so i get 30 - 50 = -20. this is negative so i think acceleration can't be negative? i am stuck with the 4 seconds but i think it is used to find the distance. it gives me two speeds, 30 mph and 50 mph. which one will be used to find the distance?

i am new here so i apologize if this was posted in the wrong place
Acceleration has a magnitude and direction. If, for example, you define acceleration in one direction to be positive, then acceleration in the opposite direction would be negative.
 
there is no tap for power. it is difficult to write the power of 2 in this site
There is, indeed, a "tap for power", it is not "difficult to write the power of 2" in this forum.
The required tap (button) is the "
X1" button; see how below...
(You must be logged in for the demonstration to play.)
Raising to a Power.gif
It is not, of course, necessary to italicize the variable (I just like to do it because it makes it look more like it would if LaTex had been used). 😉

So you could easily have written:


v = v0 + at, x = x0 + vot + ½at2, v2 = v02 + 2ax - 2ax0

though those are not the formulae I would use (at least not in that format or with those variable names). Here (and, I believe, in many other countries) we us a different nomenclature; please have a look at this post.
 
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i'm back Dr.Peterson. now i understand 50 - 30, but why we divide over time. if acceleratoin is change in velocity?

these formuals involve velocity not speed
\(\displaystyle v = v_0 + at\\x = x_0 + v_0t + 0.5at^2\\v^2 = v_0^2 + 2ax - 2ax_0\)
whenever i get speed, i treat it as velocity and i can use the formulas, correct?

i'm still not good in conversion, i use google to get this

\(\displaystyle 32 \ \text{mph} = 46 \ \text{ft/s}\)
\(\displaystyle 34 \ \text{mph} = 49 \ \text{ft/s}\)
\(\displaystyle 36 \ \text{mph} = 52 \ \text{ft/s}\)
\(\displaystyle 38 \ \text{mph} = 55 \ \text{ft/s}\)
\(\displaystyle 40 \ \text{mph} = 58 \ \text{ft/s}\)
\(\displaystyle 42 \ \text{mph} = 61 \ \text{ft/s}\)
\(\displaystyle 44 \ \text{mph} = 64 \ \text{ft/s}\)
\(\displaystyle 46 \ \text{mph} = 67 \ \text{ft/s}\)
\(\displaystyle 48 \ \text{mph} = 70 \ \text{ft/s}\)
\(\displaystyle 50 \ \text{mph} = 73 \ \text{ft/s}\)

distance = speed \(\displaystyle \times\) time \(\displaystyle = 46 \times 3.1 + 49 \times 3.3 + 52 \times 3.5 + 55 \times 3.7 + 58 \times 3.9 + 61 \times 4.1 + 64 \times 4.3 + 67 \times 4.5 + 70 \times 4.7 + 73 \times 4.9 = 2429.5 \ \text{ft}\)

how to know if the distance is correct?

i thank the explanation for writing the power of 2, but now i try to use latex to solve the problem

post 16 can you please explain acceleration is positive and negative in the same time? i'm not understand your logic!
 
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whenever i get speed, i treat it as velocity and i can use the formulas, correct?
No - that would be only "somewhat correct" in case of linear (straight-line) motion (like billiard ball moving on a table). If you are playing "catch" with a friend, most of the time the velocity of the thrown-ball is different from the speed of the ball.
 
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