Calculus test in two days!!really stuck on question an need answer asap!!please!?

Question

if the acceleration of a particle is given by a(t)=13t-3,find the velocity v(t) and the displacement s(t) given that v(0)=3 and s(0)=6

Answer 1

a = dv/dt

=> v = integral(13t-3)

= 6.5t²-3t+C

v(0)=3 => C=3

=> v= 6.5 t² - 3t + 3

v = ds/dt

=> s = integral( 6.5t²-3t+3 )

= (13/6) t³ - (3/2) t² + 3t + C

and s(0)=6 => C=6

=> s = (13/6) t³ - (3/2) t² + 3t + 6

Answer 2

acceleration is rate of change of velocity so we calculate the integral of

acceleration

a(t) = 13t-3

∫ a(t) = ∫ 13t-3 dt

v(t) = 6.5t^2-3t + 3

and velocity is rate of change of distance so calculate integral of velocity

v(t) = 6.5t^2-3t + 3

∫ v(t) = ∫ 6.5t^2-3t + 3 dt

s(t) = 13/6 t^3 - 1.5t^2 + 3t + 6

hope this helps  

Answer 3

s(t) = s(t)

v(t) = s'(t)

a(t) = s''(t)

so integrate to go backwards

if a(t) = 13t - 3

then v(t) = (13t^2)/2 -3t + c            where c = 3

and s(t) = (13t^3)/6 - (3t^2)/2 + ct + d   where d = 6

the trick is recognizing that you can make c and d whatever you want so long as the progression of derivatives from position to velocity to acceleration remains true.

Answer 4

OK, you'll need to integrate with respect to t.

v(t) = Integral(13t-3) = (13/2)xt^2 - 3t + C

v(0) = 3 = (13/2)x0^2 - 3x0 + C

=> C = 3

So v(t) = (13/2)t^2 - 3t + 3

Then integrate again to find s(t)

s(t) = Integral((13/2)t^2 - 3t + 3) = (13/6)t^3 - (3/2)t^2 + 3t + D

s(0) = 6 = (13/6)x0^3 - (3/2)x0^2 + 3x0 + D

=> D = 6

So s(t) = 13/6)t^3 - (3/2)t^2 + 3t + 6