a) -6 m/s
The particle's position is given by
[tex]x=3t^2-12t+4[/tex]
where t is the time.
The velocity of the particle can be found by calculating the derivative of the particle's position with respect to the time:
[tex]v=x'(t)=3(2t)-12=6t-12[/tex]
So, the velocity at instant t=1 can be found by substituting t=1 in this formula:
[tex]v(1 s)=6(1)-12=-6 m/s[/tex]
b) Negative direction
At instant t=1 second, the particle's velocity is
[tex]v=-6 m/s[/tex]
we see that this velocity is negative: therefore, this means that the particle is moving along the negative x-direction.
c) 6 m/s
The particle's speed is just equal to the magnitude of the velocity:
[tex]speed = |v|[/tex]
Since the speed is a scalar and velocity is a vector.
Since the velocity is
[tex]v=-6 m/s[/tex]
Then, the speed of the particle is
[tex]speed = |-6 m/s| = 6 m/s[/tex]
d) Increasing
The speed is the magnitude of the velocity, so we can write
[tex]speed = |v| = |6t-12|[/tex]
We notice that at t = 1 s, for little increments of t the term (6t) is increasing in magnitude, while the constant term -12 remains constant. This means that the whole function
[tex]|6t-12|[/tex]
is increasing at t=1 s, so the speed is increasing.
e) yes, at t = 2 s
In order to find an instant in which the velocity is zero, we just need to set v(t)=0 in the equation:
[tex]6t-12 =0[/tex]
and solving the equation for t, we find
[tex]6t=12\\t=\frac{12}{6}=2[/tex]
So, at t = 2 seconds the velocity is zero.
f) Â No
The particle is moving in the negative direction of x when the velocity vector v(t) is negative.
The velocity vector is
[tex]v(t) = 6t -12[/tex]
we see that at t=3 s, it is equal to
[tex]v(3) = 6(3)-12 = 18-12=+6[/tex]
so it is positive.
For every time t > 3 s, we notice that the term (6t) in the formula is increasing and always positive, and since it is always larger than the constant negative term (-12), we can conclude that the velocity is always positive for every time > 3 seconds.
