A swimmer loses their cap.

6 posts in this topic

Posted · Report post

As a swimmer jumps off a small bridge and begins to swim upstream, her swim cap comes off and floats downstream. Ten minutes later she turns around, swimming downstream with the same effort, past her original bridge. At the next bridge, 1000 meters away from the first, she catches the cap. What was the speed of the current? Of the swimmer?
0

Share this post


Link to post
Share on other sites

Posted · Report post

I think something is missing to calculate the swimmer's speed.



Let v=swimmer's speed, c=current, and x=time after turnaround.

iThis leads to 3 expressions.
a=10(v-c): distance the swimmer goes upstream in the first 10 minutes.
b=x(v+c): distance the swimmer goes downstream for the cap
d=(10+x)c: distance the cap floats; 1000 meters.
a+d=b
10v-10c+10c+xc=xv+xc
10v=xv
10=x
sub in:
a=10v-10c
b=10v+10c
d=20c=1000
20c=1000 yields c=50
a=10v-500
b=10v+500
10v-500+1000=10v+500, identity.

There isn't enough info to solve for v.

-1

Share this post


Link to post
Share on other sites

Posted · Report post

I think something is missing to calculate the swimmer's speed.

Let v=swimmer's speed, c=current, and x=time after turnaround.

iThis leads to 3 expressions.

a=10(v-c): distance the swimmer goes upstream in the first 10 minutes.

b=x(v+c): distance the swimmer goes downstream for the cap

d=(10+x)c: distance the cap floats; 1000 meters.

a+d=b

10v-10c+10c+xc=xv+xc

10v=xv

10=x

sub in:

a=10v-10c

b=10v+10c

d=20c=1000

20c=1000 yields c=50

a=10v-500

b=10v+500

10v-500+1000=10v+500, identity.

There isn't enough info to solve for v.

[spoiler= :thumbsup: ]

excellent job in figuring out that you cannot calculate the swimmer's speed but the current can be calculated

0

Share this post


Link to post
Share on other sites

Posted · Report post

I think something is missing to calculate the swimmer's speed.

Let v=swimmer's speed, c=current, and x=time after turnaround.

iThis leads to 3 expressions.

a=10(v-c): distance the swimmer goes upstream in the first 10 minutes.

b=x(v+c): distance the swimmer goes downstream for the cap

d=(10+x)c: distance the cap floats; 1000 meters.

a+d=b

10v-10c+10c+xc=xv+xc

10v=xv

10=x

sub in:

a=10v-10c

b=10v+10c

d=20c=1000

20c=1000 yields c=50

a=10v-500

b=10v+500

10v-500+1000=10v+500, identity.

There isn't enough info to solve for v.

[spoiler= :thumbsup: ]

excellent job in figuring out that you cannot calculate the swimmer's speed but the current can be calculated

If you give me how far upstream the swimmer got, I can get the swimmer's base speed, too! Let distance be a; v=a/10-50

0

Share this post


Link to post
Share on other sites

Posted · Report post

I think something is missing to calculate the swimmer's speed.

Let v=swimmer's speed, c=current, and x=time after turnaround.

iThis leads to 3 expressions.

a=10(v-c): distance the swimmer goes upstream in the first 10 minutes.

b=x(v+c): distance the swimmer goes downstream for the cap

d=(10+x)c: distance the cap floats; 1000 meters.

a+d=b

10v-10c+10c+xc=xv+xc

10v=xv

10=x

sub in:

a=10v-10c

b=10v+10c

d=20c=1000

20c=1000 yields c=50

a=10v-500

b=10v+500

10v-500+1000=10v+500, identity.

There isn't enough info to solve for v.

[spoiler= :thumbsup: ]

excellent job in figuring out that you cannot calculate the swimmer's speed but the current can be calculated

If you give me how far upstream the swimmer got, I can get the swimmer's base speed, too! Let distance be a; v=a/10-50

so what is the current's speed?

0

Share this post


Link to post
Share on other sites

Posted · Report post

I think something is missing to calculate the swimmer's speed.

Let v=swimmer's speed, c=current, and x=time after turnaround.

iThis leads to 3 expressions.

a=10(v-c): distance the swimmer goes upstream in the first 10 minutes.

b=x(v+c): distance the swimmer goes downstream for the cap

d=(10+x)c: distance the cap floats; 1000 meters.

a+d=b

10v-10c+10c+xc=xv+xc

10v=xv

10=x

sub in:

a=10v-10c

b=10v+10c

d=20c=1000

20c=1000 yields c=50

a=10v-500

b=10v+500

10v-500+1000=10v+500, identity.

There isn't enough info to solve for v.

[spoiler= :thumbsup: ]

excellent job in figuring out that you cannot calculate the swimmer's speed but the current can be calculated

If you give me how far upstream the swimmer got, I can get the swimmer's base speed, too! Let distance be a; v=a/10-50

so what is the current's speed?

I got the current way back in the first spoiler; c=50 meters per minute

0

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!


Register a new account

Sign in

Already have an account? Sign in here.


Sign In Now

  • Recently Browsing   0 members

    No registered users viewing this page.