1)how will the equations of the motion for an object moving with uniform velocity change?
2)show by the means of graphical method that
(a) v = u + at
(b) s = ut + 1/2at²
(c) v²-u² = 2as
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1)Pic is attached
2)a)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=OASo, v=BD + OANow,
From equation (1), OA=uSo, 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 ABor a=BD/ADBut AD=OC = t,so putting t in place of AD in the above relation, we get:
a=BD/tor 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
2b)Pic is attached.
c)Pic is attached.
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2)a)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=OASo, v=BD + OANow,
From equation (1), OA=uSo, 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 ABor a=BD/ADBut AD=OC = t,so putting t in place of AD in the above relation, we get:
a=BD/tor 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
2b)Pic is attached.
c)Pic is attached.
plz mark as brainliest
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