Answer:
v = at + u
[tex]x = ut+\frac{1}{2}at^{2}+x_{0}[/tex]
Explanation:
acceleration, a = constant
As we know that acceleration is the rate of change of velocity
[tex]a=\frac{dv}{dt}[/tex]
[tex]dv=adt[/tex]
integrate on both sides
[tex]\int dv=\int adt[/tex]
v = at + u
Where, u is the integrating constant and here it is equal to the initial velocity
Now we know that the rate of change of displacement is called velocity
[tex]v = \frac{dx}{dt}[/tex]
[tex]dx=vdt=(u+at) dt[/tex]
Integrate on both sides
[tex]\int dx=\int (u+at) dt[/tex]
[tex]x = ut+\frac{1}{2}at^{2}+x_{0}[/tex]
where, xo is the integrating constant which is initial position of the particle.