Post a New Question

physics

posted by .

a body moving with a uniform acceleration travels distances of 24m and 64m during the first two equal consecutive intervals of time each of duration 4s.determine the initial velocity and acceleration of moving body.

  • physics -

    d = vi t + (1/2) a t^2

    24 = vi (4) + (1/2) a (4)^2

    88 = vi (8) + (1/2) a (8^2)
    --------------------------------

    48 = 8 vi + 16 a
    88 = 8 vi + 32 a
    ----------------- subtract
    -40 = -16 a
    a = 2.5 m/s^2
    24 = 4 vi + 20
    vi = 1 m/s
    check my arithmetc

  • physics -

    a motor cyclist goes from station a to b at 40km/h and then returns from b to a 120km/h.calculate average speed,average velocity during the round trip

  • physics -

    case 1:
    from a to b velocity=40km/h=11.1m/s
    let 11.1=a

    case 2:
    from b to a velocity=120kh/h=33.3m/s
    let 33.3=b

    avg.speed = b-a/2
    = 33.3-11.1/2
    = 22/2
    = 11m/s
    avg. speed= 11m/s=40km/h



    avg.velocity=0
    because avg. velocity is total displacement/time.
    now the cyclist travels from a to b then from b to a, back to its original
    position. so the avg.velocity is zero.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question