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velocity and the second derivative of distance traveled with respect to time is accelera‐
tion, which is the same as the first derivative of velocity with respect to time.
Constant Acceleration
One of the simplest classes of problems in kinematics involves constant acceleration. A
good example of this sort of problem involves the acceleration due to gravity, g , on
objects moving relatively near the earth's surface, where the gravitational acceleration
is a constant 9.81 m/s 2 . Having constant acceleration makes integration over time rel‐
atively easy since you can pull the acceleration constant out of the integrand, leaving
just dt .
Integrating the relationship between velocity and acceleration described earlier when
acceleration is constant yields the following equation for instantaneous velocity:
(v1 to v2) dv = ∫ (t1 to t2) a dt
(v1 to v2) dv = a ∫ (t1 to t2) dt
v 2 - v 1 = a ∫ (t1 to t2) dt
v 2 - v 1 = a (t 2 − t 1 )
v 2 = a t 2 − a t 1 + v 1
When t 1 equals 0, you can rewrite this equation in the following form:
v 2 = a t 2 + v 1
v 2 = v 1 + a t 2
This simple equation allows you to calculate the instantaneous velocity at any given time
by knowing the elapsed time, the initial velocity, and the constant acceleration.
You can also derive an equation for velocity as a function of displacement instead of
time by considering the kinematic differential equation of motion:
v dv = a ds
Integrating both sides of this equation yields the following alternative function for in‐
stantaneous velocity:
(v1 to v2) v dv = a ∫ (s1 to s2) ds
(v 2 2 − v 1 2 ) / 2 = a (s 2 − s 1 )
v 2 2 = 2a (s 2 − s 1 ) + v 1 2
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