Formula given acceleration and distance

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Homework Help Overview

The problem involves an object being pushed across a sheet of ice, experiencing acceleration, constant speed, and deceleration. The subject area includes kinematics, specifically focusing on motion equations under uniform acceleration and deceleration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations, such as relating final and initial velocities with acceleration and time. There is an attempt to convert units and calculate speeds and distances based on given parameters.

Discussion Status

Some participants have provided calculations and interpretations of the problem, while others question the accuracy of the values used, particularly regarding time. Multiple interpretations of the problem are being explored, but there is no explicit consensus on the approach or final answers.

Contextual Notes

The original poster seeks guidance on the necessary formulas to solve the problem, indicating a need for clarification on the application of kinematic equations. There are also constraints related to significant figures in the calculations.

Pure_Anarchy
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Homework Statement


An object is pushed from rest across a sheet of ice, accelerating at 5.0 m/s over a distance of 80 cm. The object then slides at a constant speed for 4.0 s until it reaches a rough section that causes the object to stop in 2.5 s. (Assume 2 significant digits.)
(a) What is the speed of the object when it reaches the rough section?
(b) At what rate does it slow down once it reaches the rough section?
(c) What is the total distance that the object slides?

Can anyone tell me the formula I need to solve this question? Thanks in advance.

Homework Equations





The Attempt at a Solution

 
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F = ma, where F is the TOTAL force acting on the body of mass m, as a result of which the body is having an accn of 'a'.

The three formulae for uniform accn, like vf = vi + at etc.
 
Last edited:
Hi Pure Anarchy. I'm a phy I student also. I took a shot at your problem. Did you get the same answer?
5m/s = initial acceleration.
0.8m = 80cm unit conversion

Vf = Vi +at
Vf = 0 (rest) + 5m/s * 0.8m = 4m/s
4m/s for 4 seconds = 16meters

Vf in the rough will be 0
0= 4m/s + a*2.5sec
-4 = a * 2.5
-4/2.5 = a
-1.6m/s^2 = a deceleration when it hits the rough

4m to stop is -1.6m/s^2*2.5s

16M+4M = 20Meters
 
>Vf = 0 (rest) + 5m/s * 0.8m = 4m/s

Do you think you've plugged in the correct value for t?
 

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