How do you devrive equations in kinematics

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To derive the kinematic equations for constant acceleration, one can use a speed vs. time graph, where the area under the graph represents distance. The formula for distance, (delta)x = (average)v*t, is derived from the area of a rectangle or triangle formed on the graph. The average velocity, (average)v, is calculated as the mean of the initial and final velocities, leading to (average)v = (v_f + v_i)/2. Acceleration is defined as the change in velocity over time, expressed as a = (v_f - v_i)/t. Understanding these relationships allows for solving one-dimensional kinematics problems effectively.
DARTZ
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Okay, I really need to know exactly how to derive these formulas

(delta)x = (average)v*t

(average)v = v(subscript)f + v(subscript)i/2

a = v(subscript) - v(subscript)i/t


My physics teacher says you can use these 3 formulas to answer anyone dimension kinematics problem if the acceleration is constant. Now how do I derive these to get the formula. Please help I am very desperate and I need to know soon.
 
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Well distance=speed*time and acceleration is defined as the rate of change of velocity per unit time or (change in velocity)/time
 
DARTZ said:
Okay, I really need to know exactly how to derive these formulas

(delta)x = (average)v*t

(average)v = v(subscript)f + v(subscript)i/2

a = v(subscript) - v(subscript)i/tMy physics teacher says you can use these 3 formulas to answer anyone dimension kinematics problem if the acceleration is constant. Now how do I derive these to get the formula. Please help I am very desperate and I need to know soon.
These formulae are only valid if you have constant acceleration. You can derive these without calculus (well, you are using calculus but it is simple geometry) by plotting a speed vs. time diagram of an object moving with constant acceleration.

What does the area under the graph represent? (hint: area = height x width; height = v and width = \Delta t). How does the area relate to the maximum and minimum speeds? (think area of a triangle).

AM
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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