One-Dimensional Motion With Constant Acceleration

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The discussion focuses on calculating the original speed and acceleration of a truck that covers 40.0 m in 8.50 s while decelerating to a final velocity of 2.80 m/s. The correct formula for original velocity is x = 1/2(vo + v)t, leading to the original speed of 6.61 m/s, which was initially miscalculated as 12.21 m/s. For acceleration, the formula v = vo + at was used, resulting in the correct acceleration of -0.448 m/s², though the initial calculation yielded -1.1 m/s². The participant resolved their confusion regarding the calculations and expressed gratitude for the assistance received. The discussion highlights the importance of careful attention to signs in physics equations.
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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