Distance, Speed & Acceleration
Speed
Speed is defined as:
    
v = dx /dt
    where:
     v : speed in m/s
     x : distance in metres (m)
  t : time in seconds (s)
    
    
Explanation
    
        The speed of on object equals the distance travelled per unit of time.
Examples
        A helicopter is travelling at a speed of 50 m/s. How long will it take to travel
    a distance of 100 nautical miles, assuming an constant speed?
        Answer: One nautical mile equals 1,852 m. Thus, 100 nautical miles is 185,200 metres.
    The time needed follows from t = x / v, thus 185,200 / 50 =  3,704 seconds,
    or 61.7 minutes.
Acceleration
    Acceleration is defined as:
    
    
    a = dv / dt
    
    Where:
     a : acceleration in m/s
2
     v : speed in m/s
     t : time in seconds (s)
    
 
Explanation
    
        The acceleration of an object equals the change of speed per unit of time.
    
Examples
    A helicopter starts from a hover and commences a horizontal flight. In 10 seconds,
    it attains a horizontal speed of 20 m/s. What is the average acceleration?
    
    Answer: Horizontal speed in the hover is 0 m/s, thus the change of speed is 20 m/s.
    Acceleration is 20 (m/s) /10 (s) = 2 m/s
2.
    
    The helicopter now starts decelerating at a rate of 3 m/s
2.
    How long will it take it to get to a speed of zero?
    Answer: dt = dv/a = 20 / 3 = 6.67 seconds.
    
    Distance / Position
    
        When we know the speed and acceleration of an object, we can calculate the position
        at time t. We will now consider the equation involved.
    
        Position at time t:
    
        x(t) = I(v) d t
    
        ( I(t) is the notation we use for the integral over time t)
    
        Assuming a constant acceleration:
    
    
        v = v0 + at
    
        Combining: x(t) = I(v0 + at)dt= I(t) -
        I(0) = V0.t + (1/2) . at2
        + C
    
        C has the meaning of the distance at time 0. So we denote C as x0.
    
        To summarise: x(t) = x0 + v0.t
        +(1/2).at2
Explanation
    
        The position of an object at time t is defined by its default position, velocity,
        and acceleration at time t = 0 (with a constant acceleration, thus the force working
        on the object is constant).
    
        
    Example1
    A helicopter flies at a horizontal speed of 0 m/s while accelerating at a rate
    of 2 m/s
2. We define this as the starting situation (time t=0). How long will it
    take it to travel 1.6 km (about one mile)? And what is its speed after it
    has travelled this distance?
    Answer: 
    
    x(t) = x
0 + v
0.t
        +(1/2).at
2
    
    x(t) = 1600 (m)
    v
0= 0 (m/s)
    a = 2 m/s
2
    x
0=0
    
    Substitute these values to get:
    
    
    1600 = 0 + 0 . t +(1/2) . 2t
2
    t
2  = 1600
    t = 40 s.
    
    The speed is then : v=a.t = 2.40 = 80 m/s. Note that this is really rather fast (80
    m/s = 288 km/s = 180 miles/hour)!
    
    
        Example2
    Another example: A stone is falling to earth from a height of 1524 m (5000 feet).
    How long will it take it to reach the earth's surface (surface at 0 feet and no friction assumed)? Gravity acceleration is 9.8 m/s
2.
    Answer:
    
    1524 = (1/2)  9.8  t
2= 4.9
    t
2
    t
2 = 1524 / 4.9
    t = 17.6 s
 
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