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kinematics

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  • 17-05-2010 2:43pm
    #1
    rushrovers
    Registered Users Posts: 35 ✭


    can someone please help with this question... i am really stuck on it....

    a cyclist travelling at 8m/s obvserves a car, 40m distant, just beginning to accelerate away uniformly from rest. if the cyclists speed remains the same he can just catch the car.

    (i) find the acceleration of the car.

    (ii) if he had been cycling at 6m/s how close would he have got to the car?


    any help would be appreciated. cheers. :)


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    #2
    MariaBabii
    Closed Accounts Posts: 104 ✭✭MariaBabii


    (i)
    v2 = u2 + 2as
    v - speed - 8
    u - rest - 0
    a - acceleration - ?
    s - distance

    (8)2 = 0 + 2a(40)
    64 = 80a
    a = 0.8
    0.8 x 60 = 48m/s

    (ii)(6)2 = 0 + 2a(40)
    36 = 80a
    a = 0.45
    0.45 x 60 = 27m/s

    Hahaha just a guess ... i havent done applied maths in 2years lol


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    #3
    Ruski
    Registered Users Posts: 581 ✭✭✭Ruski


    MariaBabii wrote: »
    (i)
    v2 = u2 + 2as
    v - speed - 8
    u - rest - 0
    a - acceleration - ?
    s - distance

    (8)2 = 0 + 2a(40)
    64 = 80a
    a = 0.8
    0.8 x 60 = 48m/s

    (ii)(6)2 = 0 + 2a(40)
    36 = 80a
    a = 0.45
    0.45 x 60 = 27m/s

    Hahaha just a guess ... i havent done applied maths in 2years lol

    That's not right, since the cyclist travels with uniform acceleration, and doesn't stop.

    I'll have an answer out for you in ten minutes.


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    #4
    Ruski
    Registered Users Posts: 581 ✭✭✭Ruski


    Cyclist:
    u = 8m/s
    v = 8m/s
    a = 0m/s^2 (Since he travels at constant speed)
    s = 40m
    t = ?

    (i)
    Find t first:
    s=(u)(t)+1/2(a)(t)^2
    40=(8)(t) [since a=0]
    t=5s


    Use distance formula again this time for car, whose acceleration is [a]
    s=ut+1/2(at)^2
    40=1/2(a)(5)^2 [when u=0, since car starts at rest]
    40=25/2 (a)
    a=80/25
    a=3.2m/s^2

    (ii)


    How close:
    s = ut
    s = (6)(5)
    s = 30m

    He gets 10m close to the car since 40 - 30 is 10.


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    #5
    rushrovers
    Registered Users Posts: 35 ✭rushrovers


    Ruski wrote: »
    Cyclist:
    u = 8m/s
    v = 8m/s
    a = 0m/s^2 (Since he travels at constant speed)
    s = 40m
    t = ?

    (i)
    Find t first:
    s=(u)(t)+1/2(a)(t)^2
    40=(8)(t) [since a=0]
    t=5s


    Use distance formula again this time for car, whose acceleration is [a]
    s=ut+1/2(at)^2
    40=1/2(a)(5)^2 [when u=0, since car starts at rest]
    40=25/2 (a)
    a=80/25
    a=3.2m/s^2

    (ii)


    How close:
    s = ut
    s = (6)(5)
    s = 30m

    He gets 10m close to the car since 40 - 30 is 10.

    thank you so much thats great! thanks to both of yous.....


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