6. derive the equation of motion v = u + at, using graphical method.
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The body has an initial velocity u at point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration 'a' from A to B, and after time t its final velocity becomes 'v' which is equal to BC in the graph. The time t is represented by OC.
To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE.
Now, Initial velocity of the body, u=OA...... (1)And, Final velocity of the body, v=BC........ (2)But from the graph BC=BD + DCTherefore, v=BD + DC ......... (3)Again DC=OA So, v=BD + OA Now, From equation (1), OA=u So, v=BD + u ........... (4)We should find out the value of BD now. We know that the slope of a velocity – time graph is equal to acceleration, a.
Thus, Acceleration, a=slope of line AB or a=BD/AD But AD=OC = t, so putting t in place of AD in the above relation, we get:a=BD/for BD=at Now, putting this value of BD in equation (4) we get :v=at + u
This equation can be rearranged to give:v=u + at
hope it helpz
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To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE.
Now, Initial velocity of the body, u=OA...... (1)And, Final velocity of the body, v=BC........ (2)But from the graph BC=BD + DCTherefore, v=BD + DC ......... (3)Again DC=OA So, v=BD + OA Now, From equation (1), OA=u So, v=BD + u ........... (4)We should find out the value of BD now. We know that the slope of a velocity – time graph is equal to acceleration, a.
Thus, Acceleration, a=slope of line AB or a=BD/AD But AD=OC = t, so putting t in place of AD in the above relation, we get:a=BD/for BD=at Now, putting this value of BD in equation (4) we get :v=at + u
This equation can be rearranged to give:v=u + at
hope it helpz
mark as brainliest
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