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Select any two points on the graph e.g. (x(1) , y(1)) and (x(2), y(2)).

Translating 'x' & 'y' we have two points

(time(1)(s) , Velocity(1) (m/s) )& ((time(2)(s) , Velocity(2) (m/s))

Hence for Acceleration (a) = (u(1) - v(2)) /t ( difference in time ( 1 -2)s)

Substituting

a = (Velocity(1) m/s - Velocity (2) m/s)) / (time(1)s) - time(2(s))

This will give the gradient/slope, which is the acceleration

(m/s^2)

To find the line equation in the form of y = mx + b

We can take either of the above points and displace against (x,y).

Hence

(y - velocity(1) = a( x - time(1)) .

So taking an example of a car accelerating from 0 m/s (staring) to 44 m/s (30 mph) in 10 secs.

We have

a = (44m/s - - 0m/s) / (10 - 0(s) )

a = 44 m/s / 10s

a = 4.4 m/s^2

To find the line

y - 44 = 4.4 ( x - 10)

y - 44 = 4.4x - 44

y = 4.4x - 44 + 44

y = 4.4x + 0

The 'zero' indicates that the graph starts at the origin (0,0)

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lenpollock

Lvl 16
1y ago
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Wiki User

12y ago

You find the slop and y-intersept

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