POSITION VECTOR FOR PROJECTILES

POSITION VECTOR FOR PROJECTILES

Week 7

In this week regarding to use calculus for designing a roller coaster? I found some information about “Derivation of the position functions for projectiles of mass m when it launched from an initial position r_0 with an initial velocity v_0 ".

 

 

We know that the acceleration a in the time t is:

a(t) = -gj

Where:

g is the earth gravity

Then if we integrate twice:

v(t) = ∫a(t)dt = ∫-gj dt = -gtj+ _{C_1}

r(t)= ∫v(t)dt = ∫(-gtj+_{C_1})dt=-1/2 gt^2 2j+_{C_1 t+C_2}

We know that: v(0) = _{v_0}, and r(0) = r_0, to solve for the constant vectors _{C_1}  and _{C_2}, doing this produces: _{C_1}  = _{v_0},and _{C_2}  = r_0. Therefore, the position vector is:

r(t)= -1/2 gt^2 j + t_{v_0}  + r_0      Position vector

Now I know how to find the ‘Position vector”. It helps me to find the position of the costar in any time when it works.