Equations for Accelerated Motion Problem 2

AI Thread Summary
A car traveling at 60 mi/h stops in 6 seconds, leading to a calculation of the distance traveled during this time. The initial velocity was converted to meters per second, resulting in approximately 26.82 m/s. The acceleration was calculated as negative due to the car stopping, yielding about -4.47 m/s². The final distance traveled was computed to be 80.46 meters, which was confirmed as correct despite initial confusion regarding the signs of the variables. The discussion emphasized the importance of accurately identifying initial and final velocities and correctly applying the equations of motion.
Medgirl314
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Homework Statement


A car traveling at 60 mi/h stops in 6 seconds. How far does it travel during this time?

Homework Equations


a=Δv/Δt
x=xinitial+vinitialt+1/2(a)t^2


The Attempt at a Solution



First I changed units into the standard ones:
60 mi/h *1609.34 m/1 mi * 1 h/3600 s = 26.8223 (the 3 seemed to be repeating.)

Then I used a=Δv/Δt, assuming the Vinitial was zero:
26.8223/6
a≈4.47 m/s^2

Finally, I plugged my numbers into this equation:
x=xinitial+vinitialt+1/2(a)t^2

and got 80.46 m.

Would someone please check that this is correct?
 
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Medgirl314 said:

Homework Statement


A car traveling at 60 mi/h stops in 6 seconds. How far does it travel during this time?

Homework Equations


a=Δv/Δt
x=xinitial+vinitialt+1/2(a)t^2

The Attempt at a Solution



First I changed units into the standard ones:
60 mi/h *1609.34 m/1 mi * 1 h/3600 s = 26.8223 (the 3 seemed to be repeating.)

Then I used a=Δv/Δt, assuming the Vinitial was zero:
26.8223/6
a≈4.47 m/s^2

Finally, I plugged my numbers into this equation:
x=xinitial+vinitialt+1/2(a)t^2

and got 80.46 m.

Would someone please check that this is correct?
That assumption isn't correct. The given velocity in the question is the initial velocity. And it stops after 6 seconds. Acceleration would be negative. And it seems you did take that into account in the final equation, otherwise the answer wouldn't be correct. Yes, the answer is correct. But be careful for the values you choose and assumptions you make, acceleration is negative anyhow in this case.

Edit: if you take ##a## to be positive, the final equation would have a minus sign.

##x=x_{initial}+v_{initial}.t-\frac{1}{2}.a.t^2##
 
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Okay, thank you! I didn't take the negative acceleration into consideration at all, so thanks for pointing it out! I also used vintial twice when I meant to say one was vfinal. Oh! I got it backwards, the initial velocity was the given and the final velocity was zero. Thanks again!
 
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