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Physics Problem......?

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Starting from rest, a speedboat reaches a speed of 2.8 m/s in 2.7 s. What is the boat's speed after an additional 2.8 s has elapsed, assuming the boat's acceleration remains the same?

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  1. a = (Vf - Vo)/T

    where

    a = acceleration

    Vf = final velocity

    Vo = initial velocity

    T = time

    Substituting values,

    a = (2.8 - 0)/2.7

    a = 1.04 m/sec^2

    Use the same formula as above, but this time the following values are true

    Vf = final velocity

    Vo = 2.8 m/sec.

    a = 1.04 m/sec^2

    T = 2.8

    Substituting appropriate values,

    Vf - 2.8 = (1.04)(2.8)

    Vf = 2.8 + (1.04)(2.8)

    Vf = 5.7 m/sec.


  2. 5.7 m/s ?

    2.8 * (2.8 + 2.7) / 2.7 = 5.7

  3. First state all of your givens.

    Vi=0 m/s (stated by "Starting at rest" in the problem)

    Vf=2.8m/s

    t1=2.7s

    t2=5.5 (2.8+2.7)

    a=?

    We need to solve for acceleration before finding out the problem.

    Use a Kinematics Equation

    Vf=Vi+at1

    Now Isolate "a"

    Vf/a=Vi+t1

    a=Vi+t1/Vf

    a=t1/Vf

    a=2.7/2.8

    a=.9642857143 m/s^2

    Now solve for t2 using the acceleration you just solved for.

    We can use the same kinematics equation. Except t1 is now t2

    Vf=Vi+at2

    Vf=5.3 mps
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