Question

A lizard leaps 2.1\,\text m2.1m2, point, 1, start text, m, end text to the left in 0.52\,\text s0.52s0, point, 52, start text, s, end text. What was its average speed in \dfrac{\text m}{\text s} s m ​ start fraction, start text, m, end text, divided by, start text, s, end text, end fraction

Answer 1

Answer:

v = 4.03 m/s

Explanation:

It is given that,

Distance covered by a lizard, d = 2.1 m

Time taken by lizard, t = 0.52 s

We need to find the average speed of the lizard. The distance covered divided by time taken is called the average speed of lizard. So,

$v=\dfrac{d}{t}\\\\v=\dfrac{2.1\ m}{0.52\ s}\\\\v=4.03\ m/s$

So, the average speed of the lizard is 4.03 m/s.