What are the Steps for Calculating Velocity in a Coin Drop Problem?

  • Thread starter Thread starter fifaking7
  • Start date Start date
  • Tags Tags
    Rate Velocity
AI Thread Summary
To calculate velocity in a coin drop problem, the position function s(t) is derived from the equation s(t) = Vi(t) + 1/2 a t^2 + So, where a = -32 ft/s^2 and So = 800 ft. The average velocity over the interval [1,3] is calculated using the formula (V(3) - V(1)) / (3 - 1), resulting in -64 ft/sec. The instantaneous velocity at t = 3 seconds is found to be -96 ft/sec. To determine how long it takes for the coin to hit the ground, the equation s(t) = 0 must be solved, which remains unclear to the student. Understanding these steps is crucial for solving the problem effectively.
fifaking7
Messages
12
Reaction score
0

Homework Statement


A coin is dropped from a height of 800 feet. The height, s(measured in feet), at time, t(measured in seconds is given by:

s(t)= Vit + 1/2 a t^2 + So

a) find the average velocity on the interval [1,3]
b) find the instantaneous velocity when t= 3s.
c) how long does it take the coin to hit the ground?
d) find the velocity on the interval [1,3]


Homework Equations


a= -32 ft/s^2
h=800 ft


The Attempt at a Solution


a) I understand that you have to derive the position function to get the velocity function but I'm having trouble doing it. To find the average though i can do f(3)-f(1)/3-1
b) I can't find it because I'm having trouble with the derivative
c &d) I do not understand it because I missed the class that my teacher went over it when i was sick
 
Physics news on Phys.org
fifaking7 said:

Homework Statement


A coin is dropped from a height of 800 feet. The height, s(measured in feet), at time, t(measured in seconds is given by:

s(t)= Vit + 1/2 a t^2 + So

a) find the average velocity on the interval [1,3]
b) find the instantaneous velocity when t= 3s.
c) how long does it take the coin to hit the ground?
d) find the velocity on the interval [1,3]


Homework Equations


a= -32 ft/s^2
h=800 ft


The Attempt at a Solution


a) I understand that you have to derive the position function to get the velocity function but I'm having trouble doing it. To find the average though i can do f(3)-f(1)/3-1
b) I can't find it because I'm having trouble with the derivative
c &d) I do not understand it because I missed the class that my teacher went over it when i was sick

For (a): the average is not f(3)-f(1)/3-1 (which means f(3) - [f(1)/3] - 1 exactly as you have written it). If you mean (f(3)-f(1))/(3-1) you should use parentheses.

Anyway, your expression for s(t) does not have an 'h' in it; maybe you really meant to say S0 = 800. You see, these errors start to add up and can reduce your marks!

For (b): you say you are having trouble with the derivative. Please show your work, so we can see exactly what is your source of trouble.

For (c): you need to know when s(t) = 0.

Question (d) does not make much sense as written, but maybe they want you to write a formula for velocity v(t) that applies on the interval 1 ≤ t ≤ 3.

RGV
 
Letter d, i typed up b again. so letter D actually is below
D) find the velocity when the coin hits the ground.

My work so far

I used the 800 ft from the original statement as So instead of height


S(t)=Vi(0)t + 1/2 (-32ft/s^2)t^2 +So
S(t)= 0 -16t^2 + 800
v(t) = -32t

a) V(3)-V(1)/(3-1)
V(3) = -32(3)=-96
V(1) = -32(1)= -32

-96-(-32)/2 = -64ft/sec is my final answer

b) v(3)= -32(3)
v(3)= -96 ft/Sec is my final answer
C) I just don't understand
D)I know i need C but don't understand it
 
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 '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