Answer
given,
a = 2 t - 10
velocity function
we know,
[tex]\dfrac{dv}{dt}=a[/tex]
[tex]\dfrac{dv}{dt}=(2t-10)[/tex]
integrating both side
[tex]\int dv =\int (2t -10) dt[/tex]
v = t² - 10 t + C
at t = 0 v = 3
so, 3 = 0 - 0 + C
C = 3
Velocity function is equal to v = t² - 10 t + 3
Again we know,
[tex]\dfrac{dx}{dt}=v[/tex]
[tex]\dfrac{dx}{dt}=(t^2-10t + 3)[/tex]
integrating both side
[tex]\int dx =\int (t^2-10t + 3)dt[/tex]
[tex]x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t + C[/tex]
now, at t= 0 s = -4
[tex]-4 = \dfrac{0^3}{3}- 10\dfrac{0^2}{2} + 0 + C[/tex]
C = -4
So,
[tex]x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4[/tex]
Position function is equal to [tex]x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4[/tex]
We will get the equations:
Here we know the acceleration equation, so to get the motion equations we just need to integrate over time, such that the constants of integration will be the correspondent initial velocity and initial displacement.
Where:
Because you wrote the acceleration equation without units, I will also write the other two without units.
The velocity equation is:
[tex]v(t) = \int\limits {(2t - 10^2)} \, dt = (1/2)*2t^2 - 10*t + c\\\\v(t) = t^2 - 10t + 3[/tex]
Now to get the displacement equation we integrate the above one, so we get:
[tex]d(t) = \int\limits {(t^2 - 10*t + 3)} \, dt = \frac{1}{3}*t^3 - 5t^2 + 3t+ c\\\\d(t) = \frac{t^3}{3} - 5*t^2 + 3*t - 4[/tex]
If you want to learn more about motion equations, you can read:
https://brainly.com/question/19365526
