Formula given acceleration and distance

AI Thread Summary
To solve the problem, the object accelerates at 5.0 m/s² over 80 cm, reaching a speed of 4 m/s before sliding for 4 seconds at constant speed. The total distance covered during this time is 20 meters, calculated by adding the distance during acceleration and the distance traveled at constant speed. Upon hitting a rough section, the object decelerates at -1.6 m/s², taking 2.5 seconds to stop completely. The key formulas involved include Vf = Vi + at and distance calculations for both acceleration and constant speed phases.
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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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