given acceleration is 8 m/s^2, position at t= 1 is 0, at t=3 is 30, find initial position

Question

This question asks to find the initial position of a body which moves in a straight line. Its acceleration is 8 m/s^2. After 1 second it passes through the origin, after 3 seconds it is at 30 m. Find it's initial position.

This is a question in a chapter on applications of integration.

Thanks!

Answer 1

S = S₀ + V₀t + 1/2 (a) t²   this is the position formula with S₀ the initial position and V₀ the initial velocity

 

We have two time points so we can write two equations with two unknowns and solve for S₀

 

0 = S₀ + V₀ (1) + 1/2 (8) (1)²     and      30 = S₀ + V₀ (3) + 1/2 (8) (3)²

 

Simplify equations to      -4 = S₀ + V₀   and     -6 = S₀ + 3V₀

Multiply first equation by -3 and then add equations to get  

 

6 = -2S₀              Therefore initial position S₀ = -3

 

The position equation could have been derived by solving the differential equation f" = 8.

In other words, let f(t) = 8 and integrate once to get the velocity function 8t + C and integrate a second time to get the position function s(t) = 4t² + Ct + D. The constant C is the initial velocity at time zero and the constant D is the initial position at time zero. We are looking for the value of D. From here the work is the same. When t=1, s(t) = 0 and when t=3, s(t) = 30. Again you get two equations and you can solve for D.

Answer 2

at  t=1.    x=0

 

at. t=3.     x=30

 

a= 8

 

s= ut+(1/2)a*t^2.       is the formula

 

30=2u+(1/2)*8*4

 

30=2u+16

 

u=7mt/sec

 

s= ut+(1/2)*a*t^2

 

   =7+(1/2)*8*1

 

    =11

 

so, the initial position is x = -11 mt