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?

2 Answers

  • breakdown of trip between signs:

    pt 1:

    a = 2.0 m/s²

    t = 6.0 s

    d1 = 1/2at² = (0.5)(2)(6)² = 36 m

    pt 2:

    V = at = (2)(6) = 12 m/s {from data of pt 1}

    d2 = Vt = (12)(2) = 24 m

    pt 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)d3

    144 = 3d3

    d3 = 144/3 = 48 m

    Total distance between signs = d1+d2+d3 = 36+24+48 = 108 m ANS

  • “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.

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