Finding The magnitude of the velocity

[lesdayy]

Homework Statement

i am 100% just lost.. i understand (somewhat) what it is asking but i cant seem to know which equatin to use to solve this problem?

Question asked: A ball is thrown upward with an initial velocity of 14 m/s. The approximate value of g = 10 m/s2. Take the upward direction to be positive. What is the magnitude and the direction of the velocity of the ball 1 second after it is thrown?

Relevant Equations

no sure which one to even use

i have tried V=Vi - (g*t) with an answer equals to 4
as well as
d=Vi(t) + 1/2 a t^2 with an answer equals to 19

my final asnwer was 1.43 and it was still wrong
i used formula a=v/t
v=14
t=10
i divided 14/10 to get 1.4

 

[haruspex]

tried V=Vi - (g*t) with an answer equals to 4

Looks good, but you should always specify units. If this was flagged as wrong it suggests you have not entered the answer in the form expected.
It asks for "magnitude and direction", but tells you to take up as positive, which means simply answering "4m/s" should be fine. Hard to say more without seeing a screenshot of the question page and exactly what you entered.

 

[kuruman]

Which of the two equations that you used gives a velocity?
How did you get 1.43 and what kind of beast is that? What are its units?

 

[lesdayy]

Looks good, but you should always specify units. If this was flagged as wrong it suggests you have not entered the answer in the form expected.
It asks for "magnitude and direction", but tells you to take up as positive, which means simply answering "4m/s" should be fine. Hard to say more without seeing a screenshot of the question page and exactly what you entered.

 

[kuruman]

Please show your work that led you to the answer of 1.4 m/s that you entered.

 

[haruspex]

In post #1 you indicated that your first attempt was "4", not "1.4". If that was rejected, that's the screenshot I want to see.

 

[lesdayy]

Please show your work that led you to the answer of 1.4 m/s that you entered.

i used formula a=v/t
v=14
t=10
i divided 14/10 to get 1.4

 

[haruspex]

i used formula a=v/t
v=14
t=10
i divided 14/10 to get 1.4

I post #1 the formula you quoted was V=Vi - (g*t)

 

[kuruman]

You are not looking for a, the acceleration, which is given as 10 m/s². You are looking for the velocity at the end of the 1-second interval when the velocity at the beginning of the interval is 14 m/s.

 

[lesdayy]

You are not looking for a, the acceleration, which is given as 10 m/s². You are looking for the velocity at the end of the 1-second interval when the velocity at the beginning of the interval is 14 m/s.

i understand what its asking, getting to the answer is the problem. is there a formula ?

 

[haruspex]

i understand what its asking, getting to the answer is the problem. is there a formula ?

Yes! The first one you quoted in post #1!

 

[lesdayy]

Yes! The first one you quoted in post #1!

V=Vi - (g*t) ?

 

[haruspex]

V=Vi - (g*t) ?

Yes. It's the only formula you have quoted which has a velocity as an answer.

 

[lesdayy]

Yes. It's the only formula you have quoted which has a velocity as an answer.

given that, the second question changed its time from 1 second to 2 seconds which gave me an answer of -6 and was still uncorrect :(

 

[lesdayy]

i got it, thank yall VERY MUCH!

 

[haruspex]

i got it, thank yall VERY MUCH!

So what was going wrong?

 

[lesdayy]

So what was going wrong?

not knowing which formula to use!

 

[haruspex]

not knowing which formula to use!

I don't get it. In post #1 you wrote: "i have tried V=Vi - (g*t) with an answer equals to 4". Did you try 4m/s or not?

 

[kuruman]

not knowing which formula to use!

It's not that. It's understanding what the equations that you have are saying. An equation is a statement in the spoken language expressed in shorthand. For example, the equation $v=v_i+at$ replaces the following sentence in English: "The velocity at the end of a time interval $t$ is the same as the velocity at the beginning of the time interval to which one adds the acceleration multiplied by the time interval." If you can unfold the English behind the shorthand and if you know what the symbols stand for, it should be easy to figure out which equation matches the question that is asked.

 

[lesdayy]

I don't get it. In post #1 you wrote: "i have tried V=Vi - (g*t) with an answer equals to 4". Did you try 4m/s or not?

so, with the system that is used for homework.. they only allow us to check our work once. i was trying out a couple of formulasbut i didnt know which one to use. although i did get the correct answer i wasnt aware that i even did. when i submitted my answer i went with 1.4 which is why i got it wrong to begin with

 

[lesdayy]

It's not that. It's understanding what the equations that you have are saying. An equation is a statement in the spoken language expressed in shorthand. For example, the equation $v=v_i+at$ replaces the following sentence in English: "The velocity at the end of a time interval $t$ is the same as the velocity at the beginning of the time interval to which one adds the acceleration multiplied by the time interval." If you can unfold the English behind the shorthand and if you know what the symbols stand for, it should be easy to figure out which equation matches the question that is asked.

i just did a two chapters on average speed, distance, instanteneous speed, motion etc etc and i was able to understand the question asked and know whixh formula to use. for some reason something is just not clicking for this chapter, there are a few different words used to describe thesame thing and its throwing me off "(

 

[kuruman]

i just did a two chapters on average speed, distance, instanteneous speed, motion etc etc and i was able to understand the question asked and know whixh formula to use. for some reason something is just not clicking for this chapter, there are a few different words used to describe thesame thing and its throwing me off "(

Your actions belie your claims. By your admission, you obtained the wrong answer of 1.4 m/s from the equation $v=\dfrac{v_i}{g}$ which is incorrect on many grounds. You also stated in post #1 that you don't know which formula to use. That's OK, we are here to help you sort things out.

Your difficulties may very well originate in your statement "$\dots$ there are a few different words used to describe thesame thing and its throwing me off." What are the different words and what same thing are they describing? Maybe that is where your confusion with this chapter lies.

 

[nickbb12]

Whenever you have a problem like this, you should always write down what you know,
For example they gave you the initial velocity, the acceleration and time.
Write it down something like this with the correct units
u = 14ms^-1
a = -10ms^-2
t = 1s

From there if you look at motion formulae, there's one that fits perfectly to give us v, that is
v = u + at
So substitute values to give you this
v = 14 + (-10 x 1)
= 14 - 10
= 4ms^-1
I should note that ms^-1 is the exact same as m/s, just different notation.
Edit: Oh also, seeing as it's asking for direction, you could either leave it as a positive or also include the direction at the end of the statement, for example
4ms^-1 upwards
2nd Edit: You should also note that gravity will always be a downwards acceleration, hence when the ball is going up, the gravity should be having a negative impact on it. You used the displacement formula to figure out displacement, where displacement should've been done like this:
s = ut + 1/2 x a x t²
= (14 x 1) + (0.5 x (-10) x 1²)
= 14 - 5
= 9

You've done it so acceleration is positive to give you 14m.

 

[haruspex]

so, with the system that is used for homework.. they only allow us to check our work once. i was trying out a couple of formulasbut i didnt know which one to use. although i did get the correct answer i wasnt aware that i even did. when i submitted my answer i went with 1.4 which is why i got it wrong to begin with

This is one reason for keeping track of units.
"i divided 14/10 to get 1.4"
Becomes
"i divided $14m/s$ by $10m/s^2$ to get $1.4s$". Do you see that?
$\frac{\frac ms}{\frac m{s^2}}=s$.
So then you would have realized you were not getting a velocity. (You got the time until the velocity would be zero.)

But perhaps the more important point is that plugging data into random formulas is not a recipe for success. There is no point in knowing a formula unless you understand what the variables represent and in what context the formula applies.
In the case of $v=v_i-gt$, it is saying that if a vertical velocity (measured upwards) is modified by gravity over time t then it is reduced by gt.
The usual form is the more general $v_f=v_i+at$, where a is an arbitrary constant acceleration measured in the same direction as the velocities.

 

[lesdayy]

Your actions belie your claims. By your admission, you obtained the wrong answer of 1.4 m/s from the equation $v=\dfrac{v_i}{g}$ which is incorrect on many grounds. You also stated in post #1 that you don't know which formula to use. That's OK, we are here to help you sort things out.

Your difficulties may very well originate in your statement "$\dots$ there are a few different words used to describe thesame thing and its throwing me off." What are the different words and what same thing are they describing? Maybe that is where your confusion with this chapter lies.

with this specific question, its asking for the The magnitude of the velocity which in other words makes me believe that its not just asking for velocity but for something else. as well as g+10m/s^2 also being the same thing as acceleration in some formulas like Vf=Vi +at

 

[kuruman]

Velocity is a vector, it has magnitude and direction. The magnitude of the velocity is also known as "speed". For example, the magnitude of the initial velocity (or the initial speed) is 14 m/s and the direction is "up". You have to find the speed at the end of the 1-second interval and the direction which could be "up" or "down". The problem is exactly asking you to find the final velocity vector and is providing you with two boxes: one for the magnitude (or speed) and one for the direction.

 

[lesdayy]

Whenever you have a problem like this, you should always write down what you know,
For example they gave you the initial velocity, the acceleration and time.
Write it down something like this with the correct units
u = 14ms^-1
a = -10ms^-2
t = 1s

From there if you look at motion formulae, there's one that fits perfectly to give us v, that is
v = u + at
So substitute values to give you this
v = 14 + (-10 x 1)
= 14 - 10
= 4ms^-1
I should note that ms^-1 is the exact same as m/s, just different notation.
Edit: Oh also, seeing as it's asking for direction, you could either leave it as a positive or also include the direction at the end of the statement, for example
4ms^-1 upwards

can you explain why they are using g = -10 m/s2 when they are saying its g = 10 m/s2 ?

 

[nickbb12]

Gravity is ~9.8ms^-2 or in your case they simplify this to 10ms^-2
The reason gravity is negative is because of the direction of the ball. Since we are considering upwards to have a positive velocity and downwards to have a negative velocity, if gravity has a downward nature, then it should be negative. If we were to try it with positive acceleration we would have this:
u = 14ms^-1
t = 1s
a = 10ms^-2
v = u + at
= 14 + 10 x 1
= 24ms^-1
Now somethings wrong? The velocity increased upwards even though gravity was affecting it?
In order for this to be true, gravity on Earth would have to flip. So we write acceleration/gravity as negative, assuming that an upwards velocity is positive.
An easy way to understand it is that up is positive, down is negative. If gravity pulls things down then it has to be negative.

You can actually flip it around so up is negative and down is positive, like for example when skydiving you wouldn't say "My velocity is -100ms^-1, because that's wrong given the situation, in a situation where you throw a ball up, you want its upward motion to be positive and its downward motion to be negative. Hope this helped if you had any more questions let me know! :)

 

[lesdayy]

Gravity is ~9.8ms^-2 or in your case they simplify this to 10ms^-2
The reason gravity is negative is because of the direction of the ball. Since we are considering upwards to have a positive velocity and downwards to have a negative velocity, if gravity has a downward nature, then it should be negative. If we were to try it with positive acceleration we would have this:
u = 14ms^-1
t = 1s
a = 10ms^-2
v = u + at
= 14 + 10 x 1
= 24ms^-1
Now somethings wrong? The velocity increased upwards even though gravity was affecting it?
In order for this to be true, gravity on Earth would have to flip. So we write acceleration/gravity as negative, assuming that an upwards velocity is positive.
An easy way to understand it is that up is positive, down is negative. If gravity pulls things down then it has to be negative.

so when using g = 10 m/s2 the 10 would always be negative ? or depends on the direction its going ?

the second part of that question is t=2
my answer ended up being negative but the correct answer is positive

 

[haruspex]

so when using g = 10 m/s2 the 10 would always be negative ?

'g' is defined as the magnitude of the acceleration at Earth's surface, so it is always positive.
Whether g or -g is used in the equation depends what has been chosen as the positive direction. Normally one chooses the same direction as positive for all displacements, velocities, accelerations and forces. For vertical motion (on Earth's surface) up is generally chosen as positive, making the equation $v_f=v_i-gt$.

my answer ended up being negative but the correct answer is positive

It asks for magnitude and direction, filled in separately. A magnitude is never negative. For a scalar, magnitude is just the absolute value. So the form of the answer is 6m/s, downwards.

 

[kuruman]

And in the shorthand notation of physics, -6 m/s stands for "a speed of six meters per second directed downwards". If you take -6 m/s apart, the negative sign replaces "directed downwards", 6 replaces "six" and m/s replaces "meters per second".

 

[lesdayy]

And in the shorthand notation of physics, -6 m/s stands for "a speed of six meters per second directed downwards". If you take -6 m/s apart, the negative sign replaces "directed downwards", 6 replaces "six" and m/s replaces "meters per second".

how come both answers are postive when one is upward and the other is negative on the websites hw? what youre saying totally makes sense

 

[haruspex]

how come both answers are postive when one is upward and the other is negative on the websites hw? what youre saying totally makes sense

It's what "magnitude" means. The magnitude of a number is also known as its absolute value. The magnitude of 2 is 2, the magnitude of -2 is 2.
The magnitude of a velocity is the speed, which is always zero or positive and independent of direction.

 

[lesdayy]

It's what "magnitude" means. The magnitude of a number is also known as its absolute value. The magnitude of 2 is 2, the magnitude of -2 is 2.
The magnitude of a velocity is the speed, which is always zero or positive and independent of direction.

great :')
you have been very helpful! thank ypu sooooooooooo much

 

[lesdayy]

It's what "magnitude" means. The magnitude of a number is also known as its absolute value. The magnitude of 2 is 2, the magnitude of -2 is 2.
The magnitude of a velocity is the speed, which is always zero or positive and independent of direction.

one more question. when the problem given gives you "a= −10 m/s2" instead of " a= 10 m/s2" the difference is just plugging the number in as a negative ? i know the negative is just direction but as far as putting into an equation it'll be negative ?