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aliina [53]
2 years ago
12

Jaylen and ken started jogging from the same place in the opposite directions along a straight path. jaylen's speed was 30 m/min

faster than ken's speed. both of them did not change their speeds throughout the jog. at the end of the jog, they were 13.6 km apart. jaylen jogged 2.4 km more than ken. what was ken's average speed?
Mathematics
1 answer:
erica [24]2 years ago
5 0

The average speed of Ken is 70 m/min. Using the relation between the speed and distance, the required speed for Ken is calculated.

<h3>What is the relation between speed and distance?</h3>

The relationship between speed S and distance D with a period T is formed as

S = D/T

Units: m/sec (basic unit)

<h3>Calculation:</h3>

Given that,

Jaylen and Ken started jogging from the same place in opposite directions along a straight path.

This means, that the sum of distances traveled by both of them in opposite directions = 13.6 km (since it is given that at the end of their jogging, they are 13.6 km apart)

Consider,

The distance traveled by Jaylen = x km

Then the distance traveled by Ken = x - 2.4 km (since Jaylen jogged 2.4 km more than Ken)

Step 1: Finding the distance traveled by them:

x + x - 2.4 = 13.6

⇒ 2x = 13.6 + 2.4

⇒ 2x = 16

⇒ x = 8 km

So, the distance travelled by Jaylen = 8 km and by Ken = 8 - 2.4 = 5.6 km

Step 2: Finding the average speed traveled by Ken:

Since it is given that,

Jaylen's speed was 30 m/min faster than Ken's speed.

So,

For 30 m, the time = 1 min

then for 2.4 km i.e., 2400 m, the time = 80 min.

Therefore,

the average speed of Ken S = D/t

⇒ S = 5600 m / 80 min

∴ S = 70 m/min.

So, the average speed of Jaylen will be 70 m/min + 30 m/min = 100 m/min.

Thus, the average speed of Ken is 70 m/min.

Learn more about the relation between speed, distance, and time here:

brainly.com/question/3004254

#SPJ2

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