Answer:
A) Δf = - 49 m, B) v (7) = -56 m / s, a = - 8 m / s², C) t = 0.866 s
Explanation:
A) In this exercise ask to find the displacement and the average velocity, give the function of the movement
f (t) = - 4t² +3
and the range of motion 0≤ t ≤ 7
the displacement is
for t = 0
f (0) = 3
for t = 7 s
f (7) = - 4 7² +3
f (7) = -46 m
the total displacement is
Δf = f (7) - f (0)
Δf = -46 - 3
Δf = - 49 m
the average speed is defined as the displacement between the time interval
v = Df / Dt
v = -49 / 7
v = - 7 m / s
B) the speed and acceleration of the end points of the motion
the speed of defined by
v = [tex]\frac{dx}{dt}[/tex]
in this case
v = [tex]\frac{df}{dt}[/tex]
v = -8t
let's calculate
v (7) = -8 7
v (7) = -56 m / s
acceleration is defined by
a = [tex]\frac{dv}{dt}[/tex]
a = - 8 m / s²
acceleration is constant throughout the movement
C) the point where the direction changes.
This point is a point where the position goes from positive to negative, the point f = 0
0 = -4t² +3
t = √¾
t = 0.866 s
