Answer:
Explanation:
Given the position of a body modeled by the equation
S = t³+6t²-12t for 0≤t≤4 where s is in metres and t in seconds
Displacement during the interval is expressed as;
ΔS = S(4)-S(0)
s(4) = 4³+6(4)²-12(4)
S(4) = 64+96-48
S(4) = 112m
at t = 0
S(0) = 0³+6(0)²-12(0)
S(0) = 0m
The displacement of the body = 112 - 0 = 112m
average velocity is the rate of change of displacement with respect to time.
average velocity = diaplacement/time
average velocity = 112/4
average velocity = 28m/s
b) speed v = dS/dt
v(t) = 3t²+12t-12
at endpoint t = 4s
v(4) = 3(4)²+12(4)-12
v(4) = 48+48-12
v(4) = 84m/s
at t = 0
v(0) = 3(0)²+12(0)-12
v(0) = -12m/s
|v(0)| = 12m/s
acceleration = dv/dt
a(t) = 6t+12
at endpoint t = 4s
a(4) = 6(4)+12
a(4) = 24+12
a(4) = 36m/s
at t = 0
a(0) = 6(0)+12
a(0) = 0+12
a(0) = 12m/s
3) The direction of the bod is determined by the value of its velocity whether it is a negative or a positive value. A positive value shows that the object moves in the positive direction while a negative value shows that the body moves in the negative direction.
For us to determine when the body will change direction, we wil have to find the value of t at when v(t) =0
v(t) = 3t²+12t-12
3t²+12t-12 = 0
t²+4t-4 = 0
t = -4±√16-4(-4)/2
t = (-4±√32)/2
t = (-4±4√2)/2
t = -2+2√2 and -2-2√2
t = -2+2.83 and -2-2.83
t = 0.83secs and -4.83 secs
Hence the body changes velocity at t = 0.83s and -4.83secs
