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
The time at which the velocity of the particle is zero is 1 s or 3 s.
The acceleration of the particle at time, t = 5 s, is 18 m/s².
The total distance traveled by the particle at time, t = 5 s, is 23 m.
The given equation of motion;
The velocity of the particle is calculated as follows;
[tex]v = \frac{dx}{dt} = 3t^2 -12t+ 9[/tex]
The time at which the velocity of the particle is zero is calculated as follows;
[tex]3t^2 -12t+ 9 = 0\\\\t^2 -4t + 3 =0\\\\t^2-t -3t+ 3=0\\\\t(t -1) -3(t -1)=0\\\\(t-1)(t-3) =0\\\\t = 1 \ s \ \ \ or \ \ \ 3\ s[/tex]
The acceleration of the particle at time, t = 5 s, is calculated as follows;
[tex]a = \frac{dv}{dt} = 6t - 12\\\\a = 6t-12\\\\a = 6(5) - 12 \\\\a = 18 \ m/s^2[/tex]
The total distance traveled by the particle at time, t = 5 s, is calculated as follows;
[tex]x = (5)^3 - 6(5)^2 + 9(5) + 3\\\\x = 23 \ m[/tex]
Learn more about velocity of particles here: https://brainly.com/question/80295?source=archive
