Help with Problem: Find Distance Travelled in First 5.0s

  • Thread starter Thread starter Alanf718
  • Start date Start date
AI Thread Summary
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
Messages
9
Reaction score
0
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?
 
Physics news on Phys.org
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?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top