# 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 = v_{0} + at

Combining: x(t) = I(v_{0} + at)dt= I(t) -
I(0) = V_{0}.t + (1/2) . at^{2}
+ C

C has the meaning of the distance at time 0. So we denote C as x_{0}.

To summarise: x(t) = x0 + v_{0}.t
+(1/2).at^{2}

### 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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