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dv t = a = const .
dt t
Integrating dv = a dt we obtained the equation of change speed
t
t
t v t
t∫
∫ dv = a dt ⇒ v = v + a t , 2-39
0
t
t
t
0 v 0
here v is initial velocity of particle.
0
ds
Using v = a formula 2-39 can be written
t
dt
ds = v + at .
dt 0 t
Integrate last equation and assuming that initially s = s when
0
t = 0
s t t at 2
∫ ds = v dt a tdt+ t∫ ⇒ s s= 0 + v t + t , 2-40
0∫
0
0 s 0 0 2
We obtained the equation of motion with constant acceleration
of the particle.
Finally, an important differential relation involving the
displacement, velocity, and acceleration along the path may be
obtained by eliminating the time differential dt between Eqs. 2-3 and
2-6, which gives
ads = vdv 2-41
Assuming that initially s = s , v = v and integrate
0
0
2
v = v + 2(as s− 0 ) 2-42
2
t
0
The algebraic signs of s, v and a used in this equations, are
determined from the projections according vectors on the positive
direction of the axis.

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