An object has a constant acceleration of 40 ft/sec2, an initial velocity of −20 ft/sec, and an initial position of 10 ft. Find the position function, s(t), describing the motion of the object.
I'm all the way up to here, what do I do now?
a(t)=40 v(0)=-20 s(0)=10
the integral of 40 dt, antiderivative is v(t)=40t+C
40(0)+C=-20
C=-60
s(t)=integral of (40(t)-60)dt
antiderivative is 20t^2-60t+C

Respuesta :

irspow
d2x/dt2=40

dx/dt=40t+vo and vo=-20 so

dx/dt=40t-20

x(t)=40t^2/2-20t+x0 and x0=10 so

x(t)=20t^2-20t+10

Answer:

The equation that describes the motion of the object is [tex]s(t)=20t^2-20t+10[/tex]

Step-by-step explanation:

It is given that,

Acceleration of the object at time t is:

[tex]a(t)=40\ ft/s^2[/tex]

Initial velocity or velocity at t = 0, [tex]v(0)=-20\ ft/s[/tex]

Initial position or position at t = 0, [tex]s(0)=10\ ft[/tex]

Since, [tex]v(t)=\int\limits {a(t).dt}[/tex]

[tex]v(t)=\int\limits {40.dt}[/tex]

[tex]v(t)=40t+c_1[/tex]............(1)

At t = 0, [tex]v(0)=-20\ ft/s[/tex]

→ [tex]c_1=-20[/tex]

Equation (1) becomes :

[tex]v(t)=40t-20[/tex]

Since, [tex]s(t)=\int\limits {v(t).dt}[/tex]

[tex]s(t)=\int\limits {(40t-20).dt}[/tex]

[tex]s(t)=\dfrac{40t^2}{2}-20t+c_2[/tex]............(2)

At t = 0, [tex]s(0)=10\ ft/s[/tex]

→ [tex]c_1=10[/tex]

Equation (2) becomes:

[tex]s(t)=20t^2-20t+10[/tex]

Hence, this is the required solution.