anne asked in Science & MathematicsPhysics · 1 month ago

If we assume acceleration is negative downwards, so -9.8 m/s, would we have to make distance negative as well in the equation?

For example, if I am given distance, initial velocity, acceleration, and am finding time, would I have to make my given value of distance negative if I am assuming acceleration downwards is negative?

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

so like this... ?

- d = vo (t) + 1/2 (-a) t ^2

4 Answers

  • NCS
    Lv 7
    1 month ago
    Favourite answer

    General form of relevant equation:

    Δs = v₀*t + ½at²

    If your acceleration is gravity and you take "up" to be positive, then

    Δs = v₀*t - ½*9.8m/s²*t²

    and Δs will be negative (down) or positive (up) depending on the value of v₀ and t.

    But if you take "down" to be positive, then

    Δs = v₀*t + ½*9.8m/s²*t²

    The sign of Δs, v₀ and a all depend on your convention (positive is up or down).

    For your specific example, I usually take "up" to be positive and write

    Δs = v₀*t - ½*9.8m/s²*t²

    If v₀ (the vertical component of the initial velocity) is up, it's positive. If it's down, it gets a negative sign.

    If the result is negative, then the displacement is down; if positive, it's up.

  • oubaas
    Lv 7
    1 month ago

    d = Vo*t -9.8/2t^2

    if we put d = 0 we get:

    t = 2Vo/g 

    if acceleration time t' < t , the d is positive, which means you are still going forward ; if t' > t, then d becomes negative and you are getting back the starting line .

    I think an exemple will better clarify :

    Lets assume you are crossing the starting line at time to = 0 with a speed Vo = 20 m/sec and start braking with an acceleration a = -10 m/sec^2 

    after t1 = 1 sec :

    d1 = Vo*1-10/2*1^2 = 20-5 = 15 m 

    V1 = Vo+a*t1 = 20-10*1 = 10 m/sec  

    after t2 = 2 sec 

    d2 = 20*2-5*2^2 = 40-20 = 20 m

    V2 = Vo+a*t2 = 20-10*2 = 0 m/sec ... from then on the speed will change sign 

    after t3 = 4 sec 

    d3 = 20*4 -5*16 = 80-80 = 0 m (back to the starting point)

    V3 = Vo+a*t3 = 20-10*4 = -20 m/sec 

    after t4 = 5 sec 

    d4 = 20*5 -5*5^2 = 100-125 = -25 m  

    V4 = Vo+a*t4 = 20-10*5 = -30 m/sec 

  • neb
    Lv 7
    1 month ago

    The equation is a vector equation with displacement, velocity, and acceleration all being vectors. Vectors of course have both magnitude and direction. So, the direction of displacement may be in the positive or negative direction depending on the direction/magnitude of acceleration and direction/magnitude of the initial velocity. 

    Remember you can have negative acceleration but an initial positive velocity that would keep the displacement positive for awhile but at some point go negative. So, the sign of ‘d’ is determined by the RH side of the equation. It’s a good idea to make a habit of paying attention to directions and assign a consistent meaning for them when solving the problem.

  • No.  "a" is neither positive or negative.  It is simply "a."  a = -9.8, not -a.

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