Question

A paper in the journal Current Biology tells of some jellyfish-like animals that attack their prey by launching stinging cells in one of the animal kingdom's fastest movements. High-speed photography showed the cells were accelerated from rest for 700 ns at 5.30 ✕ 107 m/s2. Calculate the maximum speed reached by the cells and the distance traveled during the acceleration.

Answer 1

Explanation:

Given that,

Initial speed of cells is 0 as they were at rest

The acceleration of the cell, $a=5.3\times 10^{7}\ m/s^2$

Time, t = 700 ns

We need to find the maximum speed reached by the cells and the distance traveled during the acceleration. Let v is the final speed. So,

$v=u+at\\\\\because u =0\\\\v=at\\\\v=5.3\times 10^7\times 700\times 10^{-9}\\\\v=37.1\ m/s$

Let d is the distance traveled. Using equation of motion as follows :

$d=\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 5.3\times 10^7\times (700\times 10^{-9})^2\\\\d=1.29\times 10^{-5}\ m$

Hence, this is the required solution.