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Homework Help: I dont understand what uniform acceleration is?

  1. Sep 8, 2004 #1
    hello fellow Physicians, My names Faisal and i'm a a physics student, i'm really stuck on a question :mad: & i could really use u'r help :cool:
    An object with an initial speed of 3m/s begins to accelerate uniformly at 2m/s2. what will b its speed 5s later?

    i dont understand what uniform acceleration is?
    i really need to know the basics of physics before i can start anything advanced can u physicans recommend a good website?
    Last edited by a moderator: Sep 15, 2015
  2. jcsd
  3. Sep 8, 2004 #2


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    First of all, write down what you knwo and what you don't:
    d (distance) = ?
    vi (starting speed) = 3 m/s
    vf (final speed) = ?
    a (acceleration) = 2 m/s^2
    t (time) = 5 s

    Now, there are about 5 formulas that relate those 5 values. I can't name them off, but look at it logically: the acceleration is change of speed per second. The change of speed PER SECOND IS TWO, and it lasts FIVE SECONDS. So if you are going 2 faster every second, for 5 seconds, and you start at 3... what do you do?

    Still need more help?
  4. Sep 8, 2004 #3
    Velocity/speed is the rate of change of position.
    Acceleration measures the rate of change of velocity.
    Uniform acceleration ==>acceleration remains constant with time

    For your Problem:
    Remember these fudamental eqns:
    (i)v=u + at

    (ii)v^2=u^2 + 2as

    Try to figure these out yourself
    EDIT:where v is final velocity
    u is initial velocity
    a is acceleration
    t is time taken
    s is the displacement
    Last edited: Sep 8, 2004
  5. Sep 8, 2004 #4

    Do please give constructive answers. By just reciting formula's without defining what is what, you only make it more difficult.

  6. Sep 8, 2004 #5
    Welcome Faisal ! You should use the specific help forum next time.
    Enjoy PF !
  7. Sep 8, 2004 #6
    ''An object with an initial speed of 3m/s begins to accelerate uniformly at 2m/s2. wh

    i still dont understand the only equation which relates to this question
    is accelertaion, this is what i originally worked out 2-3/5=0.68 the answer was obviously wrong.
  8. Sep 8, 2004 #7
    Use v = v_0 + at

    v : speed
    v_0 initial speed
    a : acceleration

    so you have v = 3 + 2*5 = 13 m/s
  9. Sep 8, 2004 #8
    you have to understand what acceleration is first if you want to understand the problem properly. Dont worrry if you dont get it at first, a LOT of people do not get it untill after they work with the concept for about a month. Uniform Acceleration is the rate of change in speed, this means that every time one second passes by, the speed changes uniformly by a margin, it either decreases by that speed or increases. In your case, every second, the speed of the object increases by [tex] 2m/s^2 [/tex]. You begin at [tex] 3m/s^2 [/tex] and you travel for 5 seconds. If we break the calculation down we would see the separate speeds at each second:
    [tex]V_i = 3m/s[/tex] (no time has passed)
    [tex]V_1 = 3m/s + 2m/s [/tex] (1 second has passed)
    [tex]V_2 = 3m/s + 2m/s + 2m/s [/tex] (2 seconds have passed)
    [tex]V_3 = 3m/s + 2m/s + 2m/s + 2m/s [/tex] (3 seconds have passed)
    [tex]V_4 = 3m/s + 2m/s + 2m/s + 2m/s + 2m/s[/tex] (4 seconds have passed)
    [tex]V_5 = 3m/s + 2m/s + 2m/s + 2m/s + 2m/s + 2m/s[/tex] (5 seconds have passed)

    as you can see, the general pattern for the speed after a time [tex] t [/tex] and at constant acceleration [tex] a [/tex] and an initial speed [tex] V_o [/tex] is:
    [tex] V = V_o + at [/tex]

    if you still dont understand, try looking up some acceleration problems and try doing them, doing practice probalems is the best way to learn.
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