One-Dimensional Motion With Constant Acceleration

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SUMMARY

The discussion focuses on solving a one-dimensional motion problem involving a truck that covers 40.0 meters in 8.50 seconds while decelerating to a final velocity of 2.80 m/s. The original speed of the truck is calculated using the formula x = 1/2(vo + v)t, yielding an answer of 6.61 m/s, which aligns with the textbook solution. The acceleration is determined using the formula v = vo + at, resulting in an acceleration of -0.448 m/s². The participant initially miscalculated the original speed due to a sign error but corrected it after clarification.

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WonkySlinky
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A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final velocity of 2.80 m/s

a. Find the truck's original speed.

b. Find its acceleration


a) I'm using x=1/2(vo+v)t to find the original velocity but I keep getting vo=12.21m/s (the books answer is 6.61m/s)

b) For the second part to find acceleration I'm using v=vo+at which gives me a=-1.1m/s^2 (the books answer is -0.448m/s^2)

I don't know what I'm doing wrong
 
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[I'm using x=1/2(vo+v)t to find the original velocity but I keep getting vo=12.21m/s (the books answer is 6.61m/s)]

Show this calculation.
 
40=1/2(vo+2.8)8.5

40(2)/8.5=vo+2.8

9.41=vo+2.8

Ahhh I see now, I had a minus sign in there before. I'm getting the right answer now.

vo=6.61m/s

Thanks so much for the response, you've helped me avoid a total meltdown. Seriously, I appreciate it.
 

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