# Ball Kinematics problem

1. Sep 8, 2016

### xforeverlove21

1. The problem statement, all variables and given/known data
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

2. Relevant equations
d=vf+vi/2 * t

3. 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?

2. Sep 8, 2016

### PeroK

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

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

3. Sep 8, 2016

### superkraken

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

4. Sep 8, 2016

### PeroK

No, it isn't.

5. Sep 8, 2016

### superkraken

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

6. Sep 8, 2016

### PeroK

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

7. Sep 8, 2016

### superkraken

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

8. Sep 8, 2016

### superkraken

okay is it 50m

9. Sep 8, 2016

### PeroK

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

10. Sep 8, 2016

### superkraken

yes.i found the answer another way but cant it be solved like this?using velocity

11. Sep 8, 2016

### xforeverlove21

Yes.
But don't forget to add on the 150m

12. Sep 8, 2016

### PeroK

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 $s[0-5] = 50m$

13. Sep 8, 2016

### PeroK

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

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

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

And, the distance travelled 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 travelled 150m in the second 5s, which is 3 times the 50m it travelled 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.

14. Sep 8, 2016

### Staff: Mentor

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.

15. Sep 8, 2016

### PeroK

Sorry, I got confused about who the OP was!