# How Far Did the Ball Roll in the First 5 Seconds?

• xforeverlove21
In summary, a ball starts from rest and rolls down a hill with uniform acceleration, traveling 150 m during the second 5.0 s of its motion. To find the distance traveled during the first 5.0 s of motion, one must first recognize that the time interval given is from 5 to 10 seconds, not from 0 to 5 seconds. Therefore, the distance traveled during the first 5 seconds can be found by using the equation s = ut + (1/2)at^2, where u=0 and a=acceleration. By plugging in the values, we get s = (1/2)(1.5)(25) = 18.75
xforeverlove21

## Homework Statement

A ball starts from rest and rolls down a hill with uniform acceleration, traveling 150 m during the second 5.0 s of its motion. How far did it roll during the first 5.0 s of motion?

Vi= 0

at time t=5s -> d= 150 m

d=vf+vi/2 * t

## The Attempt at a Solution

Vi= 0 since it starts from rest

at time t=5s it travels 150 m therefore the final velocity must be 30 m/s

plug all of this into an equation

d=vf+vi/2 * t= 75 m

Total distance= 150m+ 75m= 225m

But the answer is wrong. Where did i go wrong?

xforeverlove21 said:

## Homework Statement

A ball starts from rest and rolls down a hill with uniform acceleration, traveling 150 m during the second 5.0 s of its motion. How far did it roll during the first 5.0 s of motion?

Vi= 0

at time t=5s -> d= 150 m

d=vf+vi/2 * t

## The Attempt at a Solution

Vi= 0 since it starts from rest

at time t=5s it travels 150 m therefore the final velocity must be 30 m/s

plug all of this into an equation

d=vf+vi/2 * t= 75 m

Total distance= 150m+ 75m= 225m

But the answer is wrong. Where did i go wrong?

You have misinterpreted the question. When it says "traveling 150 m during the second 5.0 s of its motion" it means that it traveled 150m from 5s to 10s after it started.

You have to find how far it rolled from 0s to 5s.

is the answer 18.75?if it is the find v at the end of 5 seconds.and substitute in the eqn s=ut+1/2at2 for 5th second to 10th

superkraken said:
is the answer 18.75?if it is the find v at the end of 5 seconds.and substitute in the eqn s=ut+1/2at2 for 5th second to 10th

No, it isn't.

what is it ?the final answer i need to find out where i made mymistake too

superkraken said:
what is it ?the final answer i need to find out where i made mymistake too

You haven't made any mistakes because you haven't made any attempt to solve the problem, as far as I can see.

okay here is my attempt
first at t=5 s
we get v=u+at2 as u=0 v=at=5a
then for time 5 to 10 seconds s=ut+1/2at2
so 150=5a x 10+1/2a x 100
150=50a+50a=100a giving a=150/100=1.5
so for 1st 5 seconds s=1/2 x 1.5 x 25=18.5

okay is it 50m

superkraken said:
okay here is my attempt
first at t=5 s
we get v=u+at2 as u=0 v=at=5a
then for time 5 to 10 seconds s=ut+1/2at2
so 150=5a x 10+1/2a x 100
150=50a+50a=100a giving a=150/100=1.5
so for 1st 5 seconds s=1/2 x 1.5 x 25=18.5

Would you like me to show you why this is wrong?

yes.i found the answer another way but can't it be solved like this?using velocity

superkraken said:
okay is it 50m
Yes.
But don't forget to add on the 150m

superkraken said:
okay here is my attempt
first at t=5 s
we get v=u+at2 as u=0 v=at=5a
then for time 5 to 10 seconds s=ut+1/2at2
so 150=5a x 10+1/2a x 100
150=50a+50a=100a giving a=150/100=1.5
so for 1st 5 seconds s=1/2 x 1.5 x 25=18.5

There are a number of problems here. You also need to be careful that "t" in these equations is the time interval. So, for 5-10s, ##t = 5## as this is an interval of 5s. It's not ##t=10##. If you were looking at motion between, say, 15 and 17 seconds, then ##t=2##, because this is a 2-second interval.

##v = u + at##, so ##v = 0 + 5a = 5a##

Here v is the speed after 5 seconds.

Now for the motion from 5-10s, ##s = ut + (1/2)at^2## becomes:

##s[5-10] = v5 + (1/2)a(5^2) = 25a +12.5a = 37.5a##

Hence ##37.5a = 150## and ##a=4ms^{-2}##

And, then, it's easy to get ##s[0-5] = 50m##

superkraken said:
yes.i found the answer another way but can't it be solved like this?using velocity

Here's what I think is the best way to do this one:

##s_1 = \frac{1}{2}at^2## This is the distance traveled from rest after ##t## seconds.

##s_2 = \frac{1}{2}a(2t)^2 = 2at^2 = 4s_1## This is the total distance traveled after ##2t## seconds.

And, the distance traveled between ##t## and ##2t## seconds is:

##s = s_2 - s_1 = 4s_1 - s_1 = 3s_1##

So, for any time ##t## an object starting from rest with uniform acceleration travels 3 times as far in the second ##t## seconds as the first ##t## seconds.

In your problem ##t = 5##. And it traveled 150m in the second 5s, which is 3 times the 50m it traveled in the first 5 seconds. But, that's just one example of a general rule for motion with uniform acceleration: 3 times as far in the second interval as in the first interval.

The question is, how much of the above conversation has helped the OP to understand and solve the problem themself? The OP hasn't posted anything since the initial question.

Helpers: Please remember that providing complete solutions to the OP is against PF rules.

gneill said:
The question is, how much of the above conversation has helped the OP to understand and solve the problem themself? The OP hasn't posted anything since the initial question.

Helpers: Please remember that providing complete solutions to the OP is against PF rules.

Sorry, I got confused about who the OP was!

## 1. What is a ball kinematics problem?

A ball kinematics problem is a physics problem that involves analyzing the motion of a ball. This includes determining the position, velocity, and acceleration of the ball at different points in time.

## 2. What are the key equations used to solve ball kinematics problems?

The key equations used to solve ball kinematics problems are the equations of motion, which include the equations for displacement, velocity, and acceleration. These equations are derived from Newton's laws of motion.

## 3. How do you approach solving a ball kinematics problem?

The first step in solving a ball kinematics problem is to identify the given information, such as the initial position and velocity of the ball. Then, use the equations of motion to calculate the unknown variables. It is important to draw a diagram and label all known and unknown quantities to help visualize the problem.

## 4. What is the difference between projectile motion and ball kinematics?

Projectile motion and ball kinematics are closely related, but there is a subtle difference between the two. Projectile motion refers to the motion of an object that is launched into the air and moves under the influence of gravity alone. Ball kinematics, on the other hand, includes the effects of air resistance and other external forces on the motion of the ball.

## 5. How does air resistance affect the motion of a ball in a kinematics problem?

Air resistance, also known as drag, affects the motion of a ball by slowing it down as it moves through the air. This means that the acceleration of the ball is not constant, and the velocity decreases over time. To account for this, the equations of motion can be modified to include a drag force term, which depends on the velocity and properties of the ball and the surrounding air.

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