Finding The magnitude of the velocity

[lesdayy]

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 ?

example "Free fall v=vi+at, a= -10 m/s2 , if up is chosen as positive direction of axis" does that mean if the quesyion is saying downward or upward use -10 or +10 ?

 

[kuruman]

If the problem specifies which direction is positive, then any magnitude of a vector that is along that direction requires a "+" sign in front of it and any magnitude of a vector that is opposite to that direction requires a "-" sign in front of it. If the problem does not specify which direction is positive, you are free to choose but you must be clear which direction is positive.

In this specific problem, you are told to take the upward direction as positive. Since up is opposite to down, and the direction of the acceleration is down, then the acceleration vector is given by $a=-g=-10~\text{m/s}^2$. The initial velocity is up and is given by $v_i=+14~\text{m/s}$. With this choice, the equation for the velocity is $$v=v_i+at=14~\text{m/s}+(-10~\text{m/s}^2)t.$$ After 1 s has elapsed, the velocity is $v=14~\text{m/s}+(-10~\text{m/s}^2)\times 1~\text{s}=4~\text{m/s}.$ The absence of the negative sign says that the velocity is still up whilst the speed has been reduced to 4 m/s.

Now consider this. Suppose the problem told you to take the down direction as positive. Then what? The equation to use is still the same, $v=v_i+at$. However, the acceleration which is down points in the positive direction, so we substitute in the equation $a=+g=+10~\text{m/s}^2.$ Likewise, the initial velocity is up which is opposite to the direction that we are told is positive. Therefore the initial velocity vector is $v_i=-14~\text{m/s}$. Putting this in the equation gives $$v=v_i+at=-14~\text{m/s}+(+10~\text{m/s}^2)t.$$ After 1 s has elapsed, the velocity is $v=-14~\text{m/s}+(+10~\text{m/s}^2)\times 1~\text{s}=-4~\text{m/s}.$ The presence of the negative sign says that the velocity is opposite to our chosen positive direction of down, i.e. it is up. Also, the speed has been reduced to 4 m/s.

The moral of the story is that, regardless of which way you choose the positive direction for the vectors and if you are careful with your substitutions, after 1 s the ball will be moving in the same direction as initially and with a reduced speed of 4 m/s. See how it works?

 

[lesdayy]

In this specific problem, you are told to take the upward direction as positive

but it said "IF"

 

[kuruman]

but it said "IF"

Where? According to your post #1

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?

 

[lesdayy]

if up is chosen as positive direction of axis

i was reffering to the question i had just asked, not the OG

 

[kuruman]

If you mean this question,

example "Free fall v=vi+at, a= -10 m/s2 , if up is chosen as positive direction of axis" does that mean if the quesyion is saying downward or upward use -10 or +10 ?

I thought I answered it in post #37. Maybe I didn't understand what you are asking. What are you asking? Please explain.

 

[haruspex]

example "Free fall v=vi+at, a= -10 m/s2 , if up is chosen as positive direction of axis" does that mean if the quesyion is saying downward or upward use -10 or +10 ?

If you choose up as positive and the acceleration is 10$m/s2$ downward then you write the equation as v=vi-10t.
What two options are you suggesting for what the question says? Please explain in more detail.

 

[lesdayy]

If you mean this question,

I thought I answered it in post #37. Maybe I didn't understand what you are asking. What are you asking? Please explain.

If you choose up as positive and the acceleration is 10$m/s2$ downward then you write the equation as v=vi-10t.
What two options are you suggesting for what the question says? Please explain in more detail.

its just another question that partains to this

its just another question that partains to og question. "Free fall. v=vi+at, a=−10m/s2, if up is chosen as positive direction of axis" it said IF... does that mean we chose + or - or ? I am not sure what the if means when given

 

[haruspex]

its just another question that partains to og question. "Free fall. v=vi+at, a=−10m/s2, if up is chosen as positive direction of axis" it said IF... does that mean we chose + or - or ?

Yes. If you choose down as positive then a=+10m/s2. But that choice reverses the signs of the velocities too, so if it is something thrown up at 5m/s then we have v=-5+10t, and a positive result means a downward velocity.