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A particle moves along x-axis and is undergoing deceleration which is proportional to the cube of its instantaneous speed v. The initial speed is v0 and after describing a path length x = x0, the speed is reduced to v_0/2. Then (a) when x = 5x0, the speed is v_0/16 (b) velocity of the particle is given by v = v_0 e^(-x/x_0 ) (c) velocity of the particle is given by v = v_0/(1+x/x_0 ) (d) when x = 3x0, the speed is v_0/5

A particle moves along x-axis and is undergoing deceleration which is proportional to the cube of its instantaneous speed v. The initial speed is v0 and after describing a path length x = x0, the speed is reduced to v_0/2. Then
    (a)    when x = 5x0, the speed is v_0/16
    (b)    velocity of the particle is given by v = v_0 e^(-x/x_0 )
    (c)    velocity of the particle is given by v = v_0/(1+x/x_0 )
    (d)    when x = 3x0, the speed is v_0/5

Grade:upto college level

1 Answers

Radhika Batra
247 Points
8 years ago

(c)

    Given a = –v3

    a = –kv3 where k is a constant of proportionality

    –v dv/dx = kv3

    Integrating ∫¦?v^(-2) dv?=-∫¦?k dx?

     –1/v = –kx + C [C is constant of integration]

    When x = 0, v = v0

    (-1)/v_0  = C

     1/v=-kx-1/v_0

     1/v-1/v_0  = kx            … (1)

    Also when x = x0, v = v_0/2

     2/v_0 -1/v_0  = kx0

     k = 1/(v_0 x_0 )

    From (1), we get v = v_0/(1+x/x_0 )

    When x = 3x0    v = v_0/4

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