Formula given acceleration and distance

In summary, an object is pushed from rest on ice, accelerating at 5m/s over a distance of 80cm. It then slides at a constant speed for 4 seconds until it reaches a rough section, where it stops in 2.5 seconds. The speed of the object when it reaches the rough section is 4m/s, with a deceleration of -1.6m/s^2. The total distance that the object slides is 20 meters. The formula used is F = ma, where F is the total force acting on the body of mass m, resulting in an acceleration of 'a'. The three formulae for uniform acceleration can also be used.
  • #1
Pure_Anarchy
1
0

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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  • #2
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:
  • #3
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
 
  • #4
>Vf = 0 (rest) + 5m/s * 0.8m = 4m/s

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

1. What is the formula for calculating acceleration given distance?

The formula for calculating acceleration given distance is a = (vf² - vi²) / (2d), where a is acceleration, vf is final velocity, vi is initial velocity, and d is distance.

2. How is this formula derived?

This formula is derived from the equation for average acceleration, which is a = Δv / Δt. By substituting the equation for velocity (vf = vi + at) into this formula and rearranging, we get the formula for acceleration given distance.

3. What are the units for each variable in this formula?

The units for acceleration (a) are m/s², the units for final velocity (vf) and initial velocity (vi) are m/s, and the units for distance (d) are meters (m).

4. Can this formula be used for any type of motion?

Yes, this formula can be used for any type of motion as long as the initial and final velocities are known and the motion is along a straight line.

5. How can I use this formula in real-world situations?

This formula can be used in various real-world situations, such as calculating the acceleration of a car during braking or the acceleration of a roller coaster on a track. It can also be used in sports science to analyze the acceleration of athletes during sprints or jumps.

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