Suppose 𝑎(𝑡) = 36𝑡 + 16, 𝑣(0) = −4, and 𝑠(1) = 15. Find 𝑠(𝑡).

Answer 1

a(t) = 36t + 16

v(t) = the integral of a(t) = 36t^2/2 +16t + c

v(0) = -4 = 18(0))^2 + 16(0) + c

c=-4

v(t) = 18t^2 + 16t -4

s(t) = the integral of v(t) = 18t^3/3 + 16t^2/2 -4t + d = (18/3)t^3 + 8t^2 - 4t + d

s(1) = 15 = 18(1)^3/3 + 8(1)^2 -4(1) + d =-6 + 8 - 4

d = 15 -6 -8 + 4 = 5

s(t) = 18t^3 + 8t^2 -4t + 5

(sort of to complete the entire series of functions, j(t) = 36

v is the change in position s

a is the change in velocity v

j is the change in acceleration a

j = jerk. sort of like driving a car and feeling the car "jerk" as acceleration change

j is the derivative of a

a is the derivative of v

v is the derivative of s

initial position at time t=0 is 5)

Answer 2

a(t) = 36t+16; Here a(t) stands for d²s/dt² which is d/dt(ds/dt) ;

Now d/dt(ds/dt) =36t +16

Integrating both sides with respect to t we get:

ds/dt = ∫(36t+16)dt ⇒ds/dt =1/2. 36t² + 16t + C, C is called integration constant.

⇒

ds/dt = 18t ² + 16t +C ; But ds/dt is the velocity V;

V(t) = 18 t² + 16t + C

Under given condition : when t = 0, v(0) = -4;⇒ -4 = 18.0 +16.0 +C ⇒C =-4

Hence, V(t) = 18t² +16t -4 But V(t) = ds/dt ⇒ds/dt = 18t² +16t -4 ;

Integrating both sides w.r.t. t we get:

∫(ds/dt) dt = ∫(18t² + 16t -4)dt

⇒ S(t) = 18.t³.1/3 + 16.t² .1/2 - 4t + C where C is the integration constant.

Applying the given initial condition S(1) =15 we get:

⇒ 15 = 18.1³.1/3 + 16.1².1/2 -4.1 +C

⇒ 15 = 6 + 8 - 4 +C ⇒C = 5

Thus S(t) = 6t³ +8t² -4t + 5 Ans.

Answer 3

a = dv/dt

a dt = dv or dv = a dt

Integrate both sides to get:

∫dv = ∫a dt

v = ∫a dt [now plug in a(t) for a to get:

v = ∫36t + 16 dt [now integrate:

v = 18t² + 16t + C₁

Using v(0) = -4, solve for C₁

-4 = 18(0)² + 16(0) + C₁ or C₁ = -4

So v(t) = 18t² + 16t - 4

Then v = ds/dt

v dt = ds or ds = v dt

∫ds = ∫v dt

s = ∫v dt

s = ∫18t² + 16t - 4 dt

s = 6t³ + 8t² - 4t + C₂

Using s(1) = 15, solve for C₂:

15 = 6(1)³ + 8(1)² - 4(1) + C₂

15 = 6 + 8 - 4 + C₂ or C₂ = 5

So s(t) = 6t³ + 8t² - 4t + 5