Thank you for that one sir...but I've a question, is it true that after finding the 12 m/s² then you must subtract it from 20 m/s² to get 8m/s² as seen in an ecz past paper app???
you can use it as well, with the coordinates (0, 0) and (10, V), and the gradient is the acceleration, a = 2 m/s^2. so you're gonna solve for V: a = (y2 - y1)/(x2 - x1) 2 = (V - 0)/( 10 - 0) 2 = V / 10 V = 20 m/s
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Thank you for that one sir...but I've a question, is it true that after finding the 12 m/s² then you must subtract it from 20 m/s² to get 8m/s² as seen in an ecz past paper app???
No it's not
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Sorry sir would like to know how we can use the formula for gradient when solving the first question
you can use it as well, with the coordinates (0, 0) and (10, V), and the gradient is the acceleration, a = 2 m/s^2.
so you're gonna solve for V:
a = (y2 - y1)/(x2 - x1)
2 = (V - 0)/( 10 - 0)
2 = V / 10
V = 20 m/s
Thanks sir
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Does it mean that verocity is same as speed
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Sir did that figure become 10t?am behind on that
Same question here
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