Question

The position of a particle moving along the x-axis varies with time according to x(t) = 5.0t2-4.0t3m. Find (a) the displacement, average velocity, average acceleration between 0.0 s and 1.0,

Answer 1

Answer:

$\Delta x=1-0=1\ m$

$\Delta v=-2-0=-2\ m.s^{-1}$

$\Delta a=-14-10=-24\ m.s^{-1}$

Explanation:

The equation governing the position of the particle moving along x-axis is given as:

$x=5\times t^2-4\times t^3$

we know that the time derivative of position gives us the velocity:

$\frac{d}{dt} x=v$

$v=10\ t-12\ t^2$

and the time derivative of of velocity gives us the acceleration:

$\frac{d}{dt} v=a$

$a=10-24\ t$

Now, when t = 0

$x=0\ m$

$v=0\ m.s^{-1}$

$a=10\ m.s^{-2}$

When t=1 s

$x_1=5\times 1^2-4\times 1^3=1\ m$

$v_1=10\times 1-12\times 1^2=-2\ m.s^{-1}$

$a_1=10-24\times 1=-14\ m.s^{-2}$

Hence,

Displacement between the stipulated time:

$\Delta x=x_1-x$

$\Delta x=1-0=1\ m$

Velocity between the stipulated time:

$\Delta v=v_1-v$

$\Delta v=-2-0=-2\ m.s^{-1}$

Acceleration between the stipulated time:

$\Delta a=a_1-a$

$\Delta a=-14-10=-24\ m.s^{-1}$

Here negative sign indicates that the vectors are in negative x direction.