Answer:
A) The initial position of the particle is 2 meters.
B) The velocity of the particle at any time t is represented by [tex]v(t) =-2\cdot t +1[/tex].
C) The acceleration of the particle at any time t is represented by [tex]a(t) = -2[/tex].
D) The velocity of the particle at the moment when [tex]s(t) = 0\,m[/tex] is -2 meters per second.
Step-by-step explanation:
A) From statement we know that [tex]s(t) = -t^{2}+t+2[/tex], where [tex]t[/tex] is the time, measured in seconds, and [tex]s(t)[/tex] is the distance, measured in meters. The initial position of the particle is calculated by evaluating the function presented above at [tex]t=0\,s[/tex]. That is:
[tex]s(0) = -(0)^{2}+(0)+2[/tex]
[tex]s(0) = 2\,m[/tex]
The initial position of the particle is 2 meters.
B) According to the Theory on Kinematics, we see that velocity is the rate of change of position in time. In other words, we need to derive the equation once:
[tex]v(t) =-2\cdot t +1[/tex] (1)
Where [tex]v[/tex] is the velocity, measured in meters per second.
The velocity of the particle at any time t is represented by [tex]v(t) =-2\cdot t +1[/tex].
C) And the acceleration is the rate of change of velocity in time. In other words, we need to derive the equation above:
[tex]a(t) = -2[/tex] (2)
The acceleration of the particle at any time t is represented by [tex]a(t) = -2[/tex].
D) At first we solve the function position for [tex]s = 0[/tex], that is:
[tex]-t^{2}+t+2=0[/tex] (3)
[tex](t-2)\cdot (t+1)=0[/tex]
The only reasonable root of the polynomial is [tex]t = 2\,s[/tex].
And now we evaluated the velocity function at given result:
[tex]v(2) = -2\cdot (2) +2[/tex]
[tex]v(2) = -2\,\frac{m}{s}[/tex]
The velocity of the particle at the moment when [tex]s(t) = 0\,m[/tex] is -2 meters per second.
