The distance between Jerry's school and his house is 1000m. He runs at a rate of...(see details)?

Question

The distance between Jerry's school and his house is 1000m. He runs at a rate of 4 meters per second, and his friend runs at a rate of 3 meters per second. If Jerry starts 5 seconds after his friend starts running, when will he catch up?

Please explain your work because I am HORRIBLE at this type of word problems.

Answer 1

His friend has a 5 second headstart. In 5 seconds he will have gone 15 meters.

Jerry runs at a rate of 4 m/s. This is 1 m/s faster than his friend.

He will catch up at a rate of 1 m/s. To catch up 15 meters will take 15 seconds. The distance to the school is irrelevant.

Using algebra:

Let T be the time that Jerry runs.
Let T+5 be the time that his friend runs:

The distance Jerry runs is:
D = 4T

The distance the friend runs is:
D = 3(T + 5)

They meet when they have run the same distance so equate these two:

4T = 3(T + 5)

Distribute the 3 through the parentheses:
4T = 3T + 15

Subtract 3T from both sides:
4T - 3T = 15
T = 15

So Jerry runs for 15 seconds to catch up.

Answer 2

Jerry's friend has a head start of 5 seconds. Convert his head start from 5 seconds to its equivalent in meters.
5s ∙ (3m/s) = 15m

After another t seconds, the friend has run t∙(3m/s) + 15m = (3t+15)m, while Jerry has run t∙(4m/s). Jerry catches up to his friend when
3t + 15 = 4t
t = 15
Jerry catches up 15 seconds after he starts running, or 20 seconds after his friend started running.

"The distance between Jerry's school and his house" was included to throw you off.

Answer 3

after 5*3/(4 -- 3) seconds = 15 seconds