Physics

A car accelerates from rest at 2.3m/s^2 for 18 seconds, then decelerates to 15 m/s over 301 meters.

A) How fast is the car moving at 18seconds? (Is it 2.3 m/s^2)

B) How far does the car travel over the entire trip? (The m/s and m/s^2 is tripping me up)

C) How long does it take for the car to travel the entire trip?

D) What is the car's acceleration in the second part of the trip?

E) What is the car's average velocity for this trip?

Thank you for anyone who can help.

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asked by Emily
  1. A.V=Vo + a*t = 0 + 2.3m/s^2*18s=41.4 m/s

    B. d = d1 + d2
    d = 0.5a*t^2 + 301 = 0.5*2.3*18^2 + 301
    = 673.6 m.

    C. V^2 = Vo^2 + 2a*d
    a = (V^2-Vo^2)/2d
    a = (15^2-41.4^2)/602 = -2.47 m/s^2 =
    deceleration.

    t2 = (V-Vo)/a = (15-41.4)/-2.47 = 10.7 s

    t1+t2 = 18 + 10.7 = 28.7 s. for the entire trip.

    D. a = -2.47 m/s^2.

    E. V = d/(t1+t2) = 673.6/28.7=23.47 m/s.

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    posted by Henry

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