Respuesta :
Answer:
[tex]\Delta x=1-0=1\ m[/tex]
[tex]\Delta v=-2-0=-2\ m.s^{-1}[/tex]
[tex]\Delta a=-14-10=-24\ m.s^{-1}[/tex]
Explanation:
The equation governing the position of the particle moving along x-axis is given as:
[tex]x=5\times t^2-4\times t^3[/tex]
we know that the time derivative of position gives us the velocity:
[tex]\frac{d}{dt} x=v[/tex]
[tex]v=10\ t-12\ t^2[/tex]
and the time derivative of of velocity gives us the acceleration:
[tex]\frac{d}{dt} v=a[/tex]
[tex]a=10-24\ t[/tex]
Now, when t = 0
[tex]x=0\ m[/tex]
[tex]v=0\ m.s^{-1}[/tex]
[tex]a=10\ m.s^{-2}[/tex]
When t=1 s
[tex]x_1=5\times 1^2-4\times 1^3=1\ m[/tex]
[tex]v_1=10\times 1-12\times 1^2=-2\ m.s^{-1}[/tex]
[tex]a_1=10-24\times 1=-14\ m.s^{-2}[/tex]
Hence,
Displacement between the stipulated time:
[tex]\Delta x=x_1-x[/tex]
[tex]\Delta x=1-0=1\ m[/tex]
Velocity between the stipulated time:
[tex]\Delta v=v_1-v[/tex]
[tex]\Delta v=-2-0=-2\ m.s^{-1}[/tex]
Acceleration between the stipulated time:
[tex]\Delta a=a_1-a[/tex]
[tex]\Delta a=-14-10=-24\ m.s^{-1}[/tex]
Here negative sign indicates that the vectors are in negative x direction.
