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

Step-by-step explanation:

Given that,

Initial speed of cells is 0 as they were at rest

The acceleration of the cell,
`a=5.3* 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* 10^7* 700* 10^(-9)\\\\v=37.1\ m/s`

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

`d=(1)/(2)at^2\\\\d=(1)/(2)* 5.3* 10^7* (700* 10^(-9))^2\\\\d=1.29* 10^(-5)\ m`

Hence, this is the required solution.