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Dutch version

One-dimensional motion

 

<v> = Δx/Δt       

<v>    average velocity  in ms-1

Δx      displacement  in m  (x final position minus x initial position)

Δt       elapsed time in s

 

Example

Someone runs 100 m in 15 s. Then he runs  40 m back in 10 s.

Find the average velocity.

 

<v> = Δx/Δt      = 60 (!) / 25 = 2.4 m/s

 

One-dimensional motion with constant velocity

 

Δx = v Δt       x – x0 = v t       s=v t      

 

One-dimensional motion with constant acceleration 

 

a = Δv/Δt                               v = V0 + at

Δv       final velocity minus initial velocity  in ms-1

a         acceleration  in ms-2

v0        initial velocity in ms-1

 

 Δx = v0t + ½at2            x = x0 + v0t + ½at2         s = v0t + ½at2                                 

 

 

 

Examples

1.         A car accelerates from rest to a velocity of 90 km/h in 15 s.

     Find the displacement during this time.

     v  = 90 km/h = 25 m/s

a = Δv/Δt = (25-0)/(15-0) = 1.67 ms-2    

s = ½ a t2 = (½)( 1.67)( 152 )= 188 m = 1.9 x102 m

 

2.         A car with a velocity of 100 km/h  is braking and stopped after 10 s

     Find the displacement during these 10 s.

     The initial velocity is 100 km/h = 27.8 m/s

     You can calculate the stopping distance in different ways

 

a.       For a movement with constant acceleration  we may write:

  <v> = (v1 + v2)/2 = (27.8 + 0)/2 = 13.9 m.    s = <v> t = (13.9)(10) = 139 m = 1.4.102 m

 

 

b.      a = Δv/Δt   =  (0 – 27.8)/10 = - 2.78 m/s2    (velocity decreases)

s = v0 t + ½ a t2 = (27.8)( 10) + (½ ) (– 2.78)( 102 )= 278 – 139 = 139 m = 1.4.102  m

                                                   

c.       The displacement is equal to the area under a velocity –time graph     ( the shaded area in the drawing )                                    

s = (½)(27.8)(10) = 139 m = 1.4.102 m 

       

v_t_diagram