Question

At time t in seconds a particles distance s(t) in micrometers from a point is given by s(t)=e^t+2-sin(t) what is the average velocity of the particle from t=1 to t=5

Answer 1

Answer:

$v_{avg}=29.499\ m.s^{-1}$

Explanation:

Given:

distance as a function of time is:

$s=e^t+2-sin\ t$

Average velocity of the particle from time t=1s to t=5s can be given by the total distance covered divided by the total time consumed.

Position at t=1s:

$s_1=e^1+2-sin\ 1$

$s_1=3.8768\ m$

Position at t=5s:

$s_5=e^5+2-sin\ 5$

$s_5=151.3721\ m$

Therefore the distance covered during this time is:

$\Delta s=s_5-s_1$

$\Delta s=151.3721-3.8768$

$\Delta s =147.4952\ m$

Now the average speed for the duration:

$v_{avg}=\frac{\Delta s}{\Delta t}$

$v_{avg}=\frac{147.4952}{5}$

$v_{avg}=29.499\ m.s^{-1}$