Help with Problem: Find Distance Travelled in First 5.0s

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The problem involves calculating the distance a ball rolls during the first 5 seconds of motion, given that it travels 150m in the second 5 seconds with uniform acceleration. The initial calculations incorrectly assumed constant velocity, leading to confusion about the initial velocity and acceleration. The correct approach involves using the displacement equation for both intervals, where the second interval's displacement can be expressed in terms of the first interval's variables. By substituting the initial velocity from the first interval into the equation, the correct distance for the first 5 seconds can be determined. Ultimately, the solution was found by replacing the initial velocity with the expression for acceleration multiplied by time.
Alanf718
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I was doing this problem, and now I wonder where I went wrong

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

This is what I did

x = 150m
t = 5.0s
with that data I got the velocity for stage 2
v = 30m/s

with that I tried to determine the acceleration so
V = Vo + at
30m/s = 0m/s + a(10s);
I got a = 3m/s

then I tried getting the x with
X = Xo + Vot + 1/2(a*t^2);
X = 0 + 0 + 1/2(3m/s*(5^2));
and I got X to be 37.5m but the book says the answer is 50m where did I go wrong?
 
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Alanf718 said:
...This is what I did

x = 150m
t = 5.0s
with that data I got the velocity for stage 2
v = 30m/s

This would be correct if there was no acceleration, i.e. if the velocity had been constant. Write down the equation of displacement for the second interval of 5 seconds first. Then see which parts of the equation you can (and need) to express with information you know about the first 5 seconds. You'll end up having only one unknown, which will be the acceleration. Then you can easily calculate the displacement in the first 5 seconds.
 
Iam ending up with 2 unknowns
writing it as
X- Xo = Vot + 1/2(a*t^2);

That leaves me with Vo unknown as well as the a, I can't solve it with 2 unkowns :-( and I don't know How I would get the initial velocity at stage two to solve this.[EDIT]
Forget it i got the answer I replace Vo with a*t in other words a*5 thanks for the help!
 
Last edited:
Alanf718 said:
Iam ending up with 2 unknowns
writing it as
X- Xo = Vot + 1/2(a*t^2);

That leaves me with Vo unknown as well as the a, I can't solve it with 2 unkowns :-( and I don't know How I would get the initial velocity at stage two to solve this.

Okay, the equation of displacement for the 'second' 5 sec is x = x0 + v0*t + 1/2*a*t^2, hence 150 = x0 + v0*5 + 1/2*a*5^2. What does xo equal? And what about v0?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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