gg asked in 科學及數學其他 - 科學 · 6 years ago

F.4 Phy mechanics

RE-EX2

MC13. ( HKCEE 2007 paper2 Q33)

The above graph shows the variation of the square of velocity v^2 with the displacement s of a particle moving along a straight line. what is the acceleration of the

particle?

(The graph:x-axis is displacement s with unit m.y-axis is square of velocity v^2 with unit m^-2 s^-2.It is a straight line join 2 points,(0,1)and(1,2).)

A.0.5 m s^-2

B.1 m s^-2

C.1.5 m s^-2

D.2 m s^-2

1 Answer

Rating
  • 土扁
    Lv 7
    6 years ago
    Favorite Answer

    Theanswer is : A. 0.5 m s⁻²

    Method 1 :

    v² = 2as + u²

    Plotting v² (y-axis) against s (x-axis), the slope

    = (2 - 1)/(1 - 0)

    = 1 m s⁻²

    Compare with the slope-intercept of st. line : y = mx + c

    slope = 2a

    2a = 1 m s⁻²

    a = 0.5 m s⁻²

    Method 2 :

    v² = 2as + u²

    When s = 0, v² = 1 :

    (1) = 2a(0) + u²

    u² = 1

    Hence, v² = 2as + 1

    When s = 1, v² = 2 :

    (2) = 2a(1) + 1

    2a = 1

    a = 0.5 m s⁻²

    Source(s): 土扁
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