Answer:
a. t=3secs and t=1sec
position is -7ft,acceleration is 18fts⁻², total distance travelled is 120ft
Explanation:
the displacement is define as
x=t³-6t²+9t+3·
since we are giving the position as a function of time, the velocity is the derivative of the position,
v=dx/dt
v=d(t³-6t²+9t+3)/dt
recall for y=axⁿ the derivative
dy/dx=a*nxⁿ⁻¹ and the derivative of a constant is zero
hence
V=3t²-12t+9
for V=0,
equivalent to t²-4t+3
solving the quadratic equation, we arrive at
(t-3)(t-1)=0
either t=3 or t=1
hence,at 3secs and 1sec the velocity is zero.
To determine the position at t=5, we substitute t=5 into
t³-6t²+9t+3
(5)³-6(5)²+9(5)+3
125-180+45+3
-7ft
The position at t=5 is -7ft
To determine the acceleration, we differentiate the velocity
a=dv/dt
a=d(3t²-12t+9)/dt
a=6t-12
at t=5
a=6(5)-12
a=18fts⁻²
Next we determine the distance covered at t=5
velocity =total distance travelled/total time taken
velocity=3t²-12t+9
V=3(25)-12(5)+9
V=24ft/s
Hence total distance travelled in t=5 is
24*5=120ft
