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How do you find a equation from a graph? - Answers
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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How do you find a equation from a graph? - Answers
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)
DuckDuckGo
How do you find a equation from a graph? - Answers
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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- og:descriptionSelect 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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