Calculate Distance Travelled in 3s with 12.5N Force on 13.5kg Cart

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The discussion focuses on calculating the distance traveled by a 13.5 kg shopping cart pushed with a force of 12.5 N over a duration of 3 seconds. The initial calculations indicated an acceleration of 0.93 m/s², leading to a final velocity of 2.79 m/s. However, the correct formula for distance, derived from the equation vf² = vi² + 2aΔx, reveals that the distance traveled is actually 4.16 meters, not 14.5 meters as initially calculated. Participants emphasized the importance of accurate formula manipulation and avoiding rounding errors in physics calculations.

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In a grocery store, you push a 13.5 kg shopping cart with a force of 12.5 N. If the cart starts at rest, how far does it move in 3.00 s?

Here's what I did:
m=13.5 kg
f=12.5 N
t=3s
v(i)=0
F=ma
a=f/m
12.5/13.5=0.93m/s^2=accel.
V(final)=0+at= 0+(.93)(3)=2.79
Vf^2=Vi^2+2a(deltax)= solve for change in X= (2a)Vf^2 (Vf = final velocity, Vi= initial velocity)
so, (2)(.93)*(2.79)^2= 14.5 m.
Ok, so what did I do wrong? I can't figure it out...
 
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Pencil said:
Vf^2=Vi^2+2a(deltax)= solve for change in X= (2a)Vf^2 (Vf = final velocity, Vi= initial velocity)
Here's your problem, you've just made a mistake on your manipulation;

[tex]v_{f}^{2} = v_{i}^{2} + 2ax \Rightarrow x = \frac{v_{f}^{2}}{2a}[/tex]

Also, becareful of inducing rounding errors.
 

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