# Calculating the distance traveled by a ball

• Joakim 1
In summary, the conversation discusses using the equation v0*t + 1/2a*t2 to find the distance traveled by a ball with an initial velocity of 3m/s upwards and an acceleration of 2.5m/s^2, after 0.4 seconds and 2.4 seconds. However, the calculated values do not match the expected results of 1 meter and 3.6 meters, respectively. The conversation also addresses the need to use parentheses in mathematical expressions to ensure correct order of operations. It is determined that the incorrect sign for the acceleration (pointing in the same direction as the velocity) is causing the discrepancies in the calculations.

## Homework Statement

The start velocity of the ball is 3m/s upward direction. Ball reaches its highest point after 1.2seconds
No air or other friction like that
i need to find how far the ball has traveled after 0.4 seconds, and after 2.4 secondsi can't get the right answers, says its supposed to be 1meter for 0.4 sec, and 3.6meter for 2.4 sec

## Homework Equations

ive tried to use v0*t + 1/2a*t2 where:
v0 = 3
t= 0.4 or 2.4
a= +/.- 2.5m/s^2

## The Attempt at a Solution

i can't get the right answers, says its supposed to be 1meter for 0.4 sec, and 3.6meter for 2.4 sec, and the equation i mentioned earlier did not work..

Joakim 1 said:

## Homework Statement

The start velocity of the ball is 3m/s upward direction. Ball reaches its highest point after 1.2seconds
No air or other friction like that
i need to find how far the ball has traveled after 0.4 seconds, and after 2.4 secondsi can't get the right answers, says its supposed to be 1meter for 0.4 sec, and 3.6meter for 2.4 sec

## Homework Equations

ive tried to use v0*t + 1/2a*t2 where:
v0 = 3
t= 0.4 or 2.4
a= +/.- 2.5m/s^2

## The Attempt at a Solution

i can't get the right answers, says its supposed to be 1meter for 0.4 sec, and 3.6meter for 2.4 sec, and the equation i mentioned earlier did not work..

Where does the acceleration ##a = \pm 2.5 m/s^2## come from? (I know the answer, but you need to get into the habit of explaining your reasoning!)

Anyway, you need to show your work. WHAT do you get for ##t = 0.4## sec? (I get 1.0 m.) What do you get for ##t = 2.4## sec? We cannot possibly help you if we cannot tell where you made your errors.

Ray Vickson said:
Where does the acceleration ##a = \pm 2.5 m/s^2## come from? (I know the answer, but you need to get into the habit of explaining your reasoning!)

Anyway, you need to show your work. WHAT do you get for ##t = 0.4## sec? (I get 1.0 m.) What do you get for ##t = 2.4## sec? We cannot possibly help you if we cannot tell where you made your errors.

Thanks for reply, and I am sorry for not beeing specific, first time posting.

What i thought i was supposed to do was to use the equation for the distance v0*t + 1/2a*t2, and put my numbers in:
3*0.4+1/2*2.5*0.4^2 i thought this would give me 1meter in answer, but i get 1.4meter

I got acceleration a=2.5m/s^2 from Δv/Δt where v is velocity and t time. v2-v1/t2-t1 = 0m-3m/s / 1,2s-0s = -3m/s / 1,2s = -2,5m/s^2

Joakim 1 said:
Thanks for reply, and I am sorry for not beeing specific, first time posting.

What i thought i was supposed to do was to use the equation for the distance v0*t + 1/2a*t2, and put my numbers in:
3*0.4+1/2*2.5*0.4^2 i thought this would give me 1meter in answer, but i get 1.4meter

I got acceleration a=2.5m/s^2 from Δv/Δt where v is velocity and t time. v2-v1/t2-t1 = 0m-3m/s / 1,2s-0s = -3m/s / 1,2s = -2,5m/s^2
OK. That makes sense for the acceleration. Even the sign makes sense.

Joakim 1 said:
What i thought i was supposed to do was to use the equation for the distance v0*t + 1/2a*t2, and put my numbers in:
3*0.4 + 1/2 * 2.5 * 0.4^2
i thought this would give me 1meter in answer, but i get 1.4meter
What sign should the acceleration have ?

Joakim 1 said:
Thanks for reply, and I am sorry for not beeing specific, first time posting.

What i thought i was supposed to do was to use the equation for the distance v0*t + 1/2a*t2, and put my numbers in:
3*0.4+1/2*2.5*0.4^2 i thought this would give me 1meter in answer, but i get 1.4meter

I got acceleration a=2.5m/s^2 from Δv/Δt where v is velocity and t time. v2-v1/t2-t1 = 0m-3m/s / 1,2s-0s = -3m/s / 1,2s = -2,5m/s^2

Please use parentheses: you wrote ##v_2 - \frac{v_1}{t_2} - t_1##, but (I hope) you meant ##\frac{v_2-v_1}{t_2-t_1}##, which you would write in plain text as (v2-v1)/(t2-t1). You need brackets because when parsing mathematical expressions, multiplication and division take precedence over addition and subtraction, so when you write v2 - v1/t2 - t1 you first compute the fraction v1/t2, then you perform subtractions. Using parentheses over-rides those priorities and gives you what you really want.

Anyway, your equation ##d = 3 t + \frac{1}{2} 2.5 \, t^2## has the acceleration pointing in the same direction as the velocity. Therefore, your ball will keep increasing its upward speed without limit, and so will blast off into outer space just like a rocket. Your ball will not reach a maximum height, it will just keep going up forever.

Last edited:
Ray Vickson said:
Please use parentheses: you wrote ##v_2 - \frac{v_1}{t_2} - t_1##, but (I hope) you meant ##\frac{v_2-v_1}{t_2-t_1}##, which you would write in plain text as (v2-v1)/(t2-t1). You need brackets because when parsing mathematical expressions, multiplication and division take precedence over addition and subtraction, so when you write v2 - v1/t2 - t1 you first compute the fraction v1/t2, then you perform subtractions. Using parentheses over-rides those priorities and gives you what you really want.

Yes that's how i ment (v2-v1) / (t2-t1).
Ray Vickson said:
Anyway, your equation ##d = 3 t + \frac{1}{2} 2.5 \, t^2## has the acceleration pointing in the same direction as the velocity. Therefore, your ball will keep increasing its upward speed without limit, and so will blast off into outer space just like a rocket. Your ball will not reach a maximum height, it will just keep going up forever.

If I do d=3*0.4+1/2*-2.5*0.4^2
where the acceleration now is negative instead, i get correct answer (1).

d=3*2.4+1/2*-2.5*2.4^2
however, when i try to use 2.4seconds instead, i get the answer 0 if i use -2.5 acceleration, and answer 14.4 if i use +2.5 acceleration, answer is supposed to be 3.6

d=3*1.2+1/2*-2.5*1.2^2
If i use 1.2 seconds for instance, then i see the ball travels 1.8meter, and knowing there is no friction in air or the surface, i can tell that for 2.4 seconds the distance traveled must have doubled from 1.8 meter (right)?

If the above is true, does it mean that the reason i get the answer 0 from d=3*2.4+1/2*-2.5*2.4^2 is because the ball travels 1.2meter in let's say this direction ->
Then when it turns, it falls back <- to its original state, so that it says 0, where the distance from start to finish is 0.
Im not sure if this is how it works, just thinking out loud

The question asks how far the ball has traveled - i.e. the total distance it has covered - not how far it now is from the starting point. (Perhaps it's a question of the interpretation of the English wording, for non-English speakers. But that's what it means.) So the correct method is to calculate how far it traveled up to the highest point, and how far it came down after that.

Joakim 1 said:
...

If the above is true, does it mean that the reason i get the answer 0 from d=3*2.4+1/2*-2.5*2.4^2 is because the ball travels 1.2meter in let's say this direction ->
Then when it turns, it falls back <- to its original state, so that it says 0, where the distance from start to finish is 0.
I'm not sure if this is how it works, just thinking out loud
If you travel 3 km to the grocery store and then 3 km back home, how far have you traveled ?

Reminds me of the joke about the man who went to a railway station and asked for a return ticket.
"Where to, sir?" asked the clerk.
"Back here, you idiot!"

Merlin3189 and SammyS
I wonder if it would help you understand, if you sketched a graph of velocity against time?
I suppose you'd have to assume that acceleration was constant, but you seem to have done that in your selection of equations.

## What is the formula for calculating the distance traveled by a ball?

The formula for calculating the distance traveled by a ball is d = v0t + (1/2)at^2, where d is the distance traveled, v0 is the initial velocity, t is the time, and a is the acceleration.

## What units should be used for calculating the distance traveled by a ball?

The units used for calculating the distance traveled by a ball depend on the units used for the initial velocity, time, and acceleration. Commonly used units include meters, seconds, and meters per second squared.

## How can air resistance affect the distance traveled by a ball?

Air resistance can affect the distance traveled by a ball by slowing it down and reducing its total distance traveled. This is because air resistance creates a force opposite to the direction of motion, causing the ball to lose velocity and travel a shorter distance.

## What factors can influence the distance traveled by a ball?

The distance traveled by a ball can be influenced by several factors, such as the initial velocity, angle of launch, air resistance, gravitational force, and surface conditions of the ball's path. Other factors such as spin, weight, and shape of the ball can also play a role.

## How can the distance traveled by a ball be measured or estimated?

The distance traveled by a ball can be measured using a ruler or tape measure, by recording the position of the ball at different time intervals, or by using specialized equipment such as a radar gun or motion sensor. It can also be estimated by using mathematical calculations based on the ball's initial velocity and acceleration.