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

[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

Homework Equations

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?

 

[PeroK]

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

Homework Equations

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.

 

[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

 

[PeroK]

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.

 

[superkraken]

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

 

[PeroK]

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.

 

[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

 

[superkraken]

okay is it 50m

 

[PeroK]

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?

 

[superkraken]

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

 

[xforeverlove21]

okay is it 50m

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

 

[PeroK]

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 $s[0-5] = 50m$

 

[PeroK]

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.

 

[gneill]

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.

 

[PeroK]

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!