Record the speed of movement of the animals in descending order: swallow 17.5 m / s; snail 1.4 mm / s

Record the speed of movement of the animals in descending order: swallow 17.5 m / s; snail 1.4 mm / s; shark 500 m / min; hare 60 km / h. How far will the swallow fly in 30 minutes?

Given:

v1 = 17.5 meters per second – the speed of the swallow;

v2 = 1.4 millimeters per second – the speed of the snail;

v3 = 500 meters per minute – shark speed;

v4 = 60 kilometers per hour – the speed of the hare;

t = 30 minutes = 1800 seconds – time span.

It is required to write down the speeds of animals in descending order, and also to determine S (meter) – the path that the swallow will fly in time t.

Let’s bring the speed of movement of animals to a single system of units: meter per second.

v2 = 1.4 mm / s = 1.4 / 1000 = 0.0014 m / s;

v3 = 500 m / min = 500/60 = 8.3 m / s;

v4 = 60 km / h = 60 * 10/36 = 16.7 m / s.

Then, in decreasing order of speed, they will be written like this:

17.7> 16.7> 8.3> 0.0014 (swallow, hare, shark, snail).

The path traversed by the swallow will be equal to:

S = v1 * t = 17.5 * 1800 = 31,500 meters = 31.5 kilometers.

Answer: speeds in descending order: swallow, hare, shark, snail. In half an hour, the swallow will fly a path equal to 31.5 kilometers.