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# people that are good at physics! how do you do this problem?

John walks 1.20 km north, then turns right and walks 1.00 km east. His speed is 2.00 m/s during the entire stroll.

A) What is the magnitude of his displacement, from beginning to end?

d= ? km

B) If Jane starts at the same time and place as John, but walks in a straight line to the endpoint of John's stroll, at what speed should she walk if she wants to arrive at the endpoint just when John does?

v= ? m/s

can answer either part A or B or both

i still don't get how to do the second part :(

### 2 Answers

- Al PLv 71 decade agoFavorite Answer
The distance John walks

s = 1.2+1 = 2.2 km = 2200 m

John's Time:

t = s/v = 2200/2 = 1100 s

A)

The magnitude of his displacement:

D = Sqrt(1.2^2+1^2) = 1.562 km = 1562 m

B)

The speed Jane needs during John's time:

v = D/t = 1562/1100 = 1.42 m/s

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- 1 decade ago
This is not a real physics question,

((1.2km)^2 + (1.0)^2 ) )^(1/2) is the displacement.

The second question requires some physics:

Divide the the displacement by the the distance from Jane start to John's end place.

Then you will have the distance, the time for John is the distance he went and then the time required for Jane.

You will have the answer.

Physics is fun and simple math works too.

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