Fundamental theorem of calculus and its application to 1D kinematics

1) Substitute the velocity vx(t) from formula (II) into formula (I) and write out the resulting formula for Delta(x). The displacement Delta(x) can be calculated from the knowledge of the initial velocity vix and the acceleration function ax(s). You should find the term in vix separated from a double integral term in the acceleration.

2. Using the previous formula with the double integral of ax(s), show that for a constant acceleration ax, the double integral becomes the familiar (1/2)ax(tf-ti)^2 formula you have seen in kinematics in PHYS120.

LET Q( ) = X /Y ( QUANTITY = POSITION ) THEN Q'(( = X'(t = duly = NY (t ) [ RATE- OF-CHANGE OF POSITION WITH E EXPRESSES HOW FAST THE POSITION IS CHANGING WITH TIME , WHICH WE CALL VELOCITY NX DISPLACEMENT ( OF AN OBJECT BETWEEN t ; AND t ) NOTE: INTEGRAL OF THE INSTANTANEOUS VELOCITY BETWEEN ti ANDY VELOCITY + CHANGE IN POSITION NX $ DX j N , ( Hat : Ax = x/ . x ; ( I ) ACCELERATION # CHANGE IN VELOCITY ax $ DUX BUT RATE AT WHICH THEN Q(4 = NYCH = dx( x) = all) _ RATE - OF - CHANGE OF NX WITH t. IN OTHER words, THE CHANGE HAPPENS MOW FOST N CHANCES WITHE WHICH IS Ayo CALLED' ACCELERATION a a x ( + ) dt = Dux - NJ x - Nix ( II )a x ( +) dt : Dux - NJ x - Nix ( II ) Jaxls jobs = Nx(t ) - Nix