Part II: s(t), v(t) and a(t) : position vector, velocity vector and acceleration vector related Def: Instantaneous velocity (speed) (1) V(1) = s '(t) =" Instantaneous acceleration: (2) a(t) = v '(t) = (3) a(t)=s"(t) Thus: (4) s(t)=S v(t) dt v(t) = S a(t) dt Your Problem: is to prove (4) i) Given a(t) = -g g = gravity downward Find an expression for v(t) v(t) = ii) Given an initial condition of : v (0) = vo, solve for C and find general equation for v(t): i) With your expression for v(t), find s(t). Then use: s(0) = so to find constant C and general equation for s(t): s(t)
Part II: s(t), v(t) and a(t) : position vector, velocity vector and acceleration vector related Def: Instantaneous velocity (speed) (1) V(1) = s '(t) =" Instantaneous acceleration: (2) a(t) = v '(t) = (3) a(t)=s"(t) Thus: (4) s(t)=S v(t) dt v(t) = S a(t) dt Your Problem: is to prove (4) i) Given a(t) = -g g = gravity downward Find an expression for v(t) v(t) = ii) Given an initial condition of : v (0) = vo, solve for C and find general equation for v(t): i) With your expression for v(t), find s(t). Then use: s(0) = so to find constant C and general equation for s(t): s(t)
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Transcribed Image Text:b) Check your result by first integrating and then differentiating f(x) =(t® +1)
Part II: s(t), v(t) and a(t): position vector, velocity vector and acceleration vector related
Def: Instantaneous velocity (speed)
(1) V(t) = s '(t) ="
dt
Instantaneous acceleration:
dv
(2) a(t) = v '(t) =
(3) a(t) =s"(t)
Thus:
(4) s(t)= S v(t) dt
v(t) = S a(t) dt
Your Problem: is to prove (4)
i)
Given a(t) = -g
g = gravity downward
Find an expression for v(t)
v(t) =
ii)
Given an initial condition of : v (0) = vo, solve for C and find general equation for v(t):
ili)
With your expression for v(t), find s(t). Then use: s(0) = so to find constant C and general equation
for s(t):
s(t) =
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