Simplifying the integral results in the equation v(t) = -9.8t C_1, where C_1 is the initial velocity (in physics, this the initial velocity is v_0). We can use this knowledge (and our knowledge of integrals) to derive the kinematics equations.įirst, we need to establish that acceleration is represented by the equation a(t) = -9.8.īecause velocity is the antiderivative of acceleration, that means that v'(t) = a(t) and v(t) = int. We know that acceleration is approximately -9.8 m/s^2 (we're just going to use -9.8 so the math is easier) and we know that acceleration is the derivative of velocity, which is the derivative of position. We usually start with acceleration to derive the kinematic equations. 3 m/s ) 2 − 4 t, equals, start fraction, minus, 18, point, 3, start text, space, m, slash, s, end text, plus minus, square root of, left parenthesis, 18, point, 3, start text, space, m, slash, s, end text, right parenthesis, squared, minus, 4, open bracket, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, left parenthesis, minus, 12, point, 2, start text, space, m, end text, right parenthesis, close bracket, end square root, divided by, 2, open bracket, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, close bracket, end fraction For instance, say we knew a book on the ground was kicked forward with an initial velocity of v 0 = 5 m/s v_0=5\text t = 2 − 1 8.
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