Q.4.21 A particle starts from the origin at = 0 s with a velocity of 10.0 ? m/s and moves in the x-y plane with a constant acceleration of (8.0 ? + 2.0 ?) m s-2.

(a) At what time is the x- coordinates of the particle 16 m? What is the y-coordinate of the particle at that time?

 

(b) What is the speed of the particle at the time?

0 9 Views | Posted 4 months ago
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    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    Ans.4.21:

    (a) Velocity , v? = 10.0 ? m/s

    Acceleration, a? = (8.0 ? + 2.0 ?) m s-2

    We know a? = dv?dt = 8.0 ? + 2.0 ?

    dv? = (8.0 ? + 2.0 ?)dt

    Integrating both sides we get v? (t) = 8.0t ? + 2.0t ? + u? ,

    Where, u?= velocity vector of the particle at t =0

    v?= velocity vector of the particle at time t

    But v? = ?drdt

    dr? = v? dt

    = (8.0t ? + 2.0t ? + u? )dt

    Integrating both sides with the condition at t = 0, r =0 and at t =t, r = r

    r?= u? t + ½ * 8.0 t2 ? + ½ * 2.0 * t2 ? = u? t + 4.0 t2 ? + t2 ?

    Substituting the value of u? , we get

    r?= ( 10.0 ?)t + 4.0 t2 ? + t2 ? . This equation can be expressed as

    x ? + y ? = 4.0 t2 ? + ( 10.0t + t2) ?

    Since the motion of the particle is confined to the x-y plane, on equating the coefficients of ? and ?, we get

    x = 4.0 t2 and y = 10.0t + t2

    t = x/4

    (a) When x = 16m, t = 2 s, y = 24m

    (b) Velocity of the particle

    v? (t) = 8.0t ? + 2.0t ? + u?

    At t = 2 s,

    v? (t) = 8.0 * 2 ? + 2.0 *2 ? + 10 * ? = 16 ? + 14 ?

    The magnitude of v? (t) is given by

    v? = ( 162 + 142)1/2 = 21.26 m/s

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