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Kelsie Sawayn

Feb 20, 2025

A car starts from rest at a stop sign. It accelerates at 2.0 m/s^2 for 6.0 seconds, coasts for 2.0?

A car starts from rest at a stop sign. It accelerates at 2.0 m/s^2 for 6.0 seconds, coasts for 2.0s, and then slows down at a rate of 1.5 m/s^2 for the next stop sign.How far apart are the stop signs?
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Madie Hilpert

Feb 20, 2025

breakdown of trip between signs: pt 1:a = 2.0 m/s² t = 6.0 sd1 = 1/2at² = (0.5)(2)(6)² = 36 mpt 2:V = at = (2)(6) = 12 m/s {from data of pt 1}d2 = Vt = (12)(2) = 24 mpt 3: a = – 1.5 m/s²(Vf)² = (Vi)² + 2a(d3) {where Vf = 0, Vi = 12, a = -1.5}0 = (12)² – (2)(1.5)d3144 = 3d3d3 = 144/3 = 48 mTotal distance between signs = d1+d2+d3 = 36+24+48 = 108 m ANS... Show More
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Madie Hilpert

Feb 20, 2025

“starts off from relax and speeds up at a velocity of three m/s^2 for 20 seconds” exchange in v = at. Use the given values of a and t to ascertain the excellent velocity. “decelerates on the value of 5 m/s^2” exchange in v = at. Use the given values of (exchange in v) and a to ascertain t.... Show More
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