Question

A biologist looking through a microscope sees a bacterium at r⃗ 1=2.2i^+3.7j^−1.2k^μm(1μm=10−6m). After 6.2 s , it's at r⃗ 2=4.6i^+1.9k^μm. find average speed

Answer 1

The Average velocity for the bacterium is 0.75 unit/sec.

Explanation:

The given values are in the vector form

Where,

dS = distance covered

dT = time interval

Now, to calculate distance covered, we have

$|d S|=\sqrt{d S^{2}}$

&

$d S=r_{2}-r_{1}$

d S=(4.6 i+1.9 k)-(2.2 i+3.7 j - 1.2 k)

d S=(4.6-2.2) i+(0-3.7) j+(1.9+1.2) k

d S=2.4 i-3.7 j+3.1 k

Now, putting these values in the standard formula to evaluate the average velocity, we get;

$v_{a v g}=\frac{|\mathrm{d} S|}{d T}$

$v(a v g)=\frac{|\sqrt{\left\{\left(2.4^{2}\right)+\left(3.7^{2}\right)+\left(3.1^{2}\right)\right\}}|}{7.2}$

As dT=7.2 sec

Now,

Solving the equation, we get;

$v(a v g)=\frac{5.390732789}{7.2}$

$\begin{aligned}&v(a v g)=\frac{5.39}{7.2}\\&v(a v g)=0.748611111\\&v(a v g)=0.75 \text { units / sec }\end{aligned}$

Hence, the average velocity for the bacterium is 0.75 unit/sec.