Grade 12 Kinematics question

In summary, a sunbather is drifting downstream on a raft and jumps off to swim against the current for 15 minutes. After turning around and swimming downstream again, she catches up to the raft 1.0km downstream from the bridge. Using the variables u for her speed with respect to the water, v for the speed of the water in the river, t for the time of the downstream swim, and d for the distance swum upstream, it can be solved that t = .25 and v = 2 with any value for u.
  • #1
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Homework Statement


A sunbather, drifting downstream on a raft, drives off the raft just as it passes under a bridge and swims against the current for 15min. She then turns and swims downstream, making the same total effort and overtaking the raft when it is 1.0km down stream from the bridge. What is the speed of the current in the river?

t= 0.25h
dr=1.0km

Homework Equations





The Attempt at a Solution



This question is owning me. Hard. I've tried at least 10different methods to crack this but, alas, I figure afterwards that my logic must be wrong, or stuff doesn't cancel out... I'll try to show my work at best I can but most of it is abstract so... Here goes.

Ok so we know that she travels x distance in 15mins opposite to the raft, so her velocity would be.

Vgirl.ground= -(x/0.25)

We also know that in the 15mins the raft travels y distance, and is effectively traveling with the current so,

Vraft= y/0.25

Now I know that the velocity that the girl is going at is her velocity relative the the ground, so we need to make her velocity relative to the water.
Vgirl.water= Vgirl.ground - Vraft

After around here my logic is hazy and I'm not quite sure how to continue next... Or if even what I'm doing up to this point is correct. Help would be appreciated =)
 
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  • #2
Sorry, I am old and can't wrap my mind around the particular variables you have used, but I see that you are on the right track. Humor me and think of
u = speed of swimmer with respect to the water
v = speed of water in river
t = time of swim downstream
t + .25 = time raft drifts downstream
d = distance swum upstream

Then write a d = vt equation for the raft, the upstream swim and the downstream swim.
You have unknowns u, v, t and d. And only 3 equations.
1 = v(t + .25) for the raft
d = (u-v)*.25 for swimming upstream

Curious - I find a solution of t = .25, v = 2 with ANY value you like for u.
 
  • #3


Dear student,

Thank you for presenting your attempt at solving this kinematics problem. I understand that you have been struggling with finding the solution and have tried multiple methods to solve it. I will do my best to guide you through the problem and provide a solution.

First, let's define some variables to make it easier to understand the problem:

- Vg: velocity of the girl swimming upstream
- Vr: velocity of the raft downstream
- Vc: velocity of the current
- t: time taken by the girl to swim upstream (15 minutes or 0.25 hours)
- dr: distance between the bridge and the point where the girl overtakes the raft (1.0 km)

Now, let's look at the situation when the girl is swimming upstream against the current. We can use the formula: distance = velocity x time (d = vt) to calculate the distance traveled by the girl and the raft.

For the girl, the distance traveled upstream is given by d = Vg x t. Since we know that the girl and the raft are making the same effort, we can say that the distance traveled by the raft downstream is also d = Vr x t.

Since the girl overtakes the raft after both of them have traveled for 15 minutes, we can equate the two distances and solve for Vg and Vr:

Vg x t = Vr x t

Vg = Vr

This means that the girl's velocity when swimming upstream is the same as the raft's velocity when traveling downstream.

Now, let's look at the situation when the girl turns and swims downstream. We know that it takes the girl the same amount of time (15 minutes or 0.25 hours) to swim downstream and overtake the raft. Using the same formula (d = vt), we can calculate the distance traveled by the girl and the raft.

For the girl, the distance traveled downstream is given by d = (Vg + Vc) x t. Since the girl is now swimming with the current, we add the velocity of the current (Vc) to her velocity (Vg).

For the raft, the distance traveled downstream is given by d = Vr x t.

Since we know that the girl overtakes the raft after both of them have traveled for 15 minutes, we can equate the two distances and solve for Vc:

(Vg + Vc) x t = Vr x t
 

1. What is Kinematics?

Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion. It focuses on describing the position, velocity, and acceleration of an object over time.

2. What is Grade 12 Kinematics?

Grade 12 Kinematics is a specific topic within the field of Kinematics that is typically taught in high school physics courses. It covers the concepts of displacement, velocity, acceleration, and motion graphs.

3. How is Kinematics related to other branches of physics?

Kinematics is closely related to other branches of physics, such as dynamics and mechanics. Dynamics deals with the forces that cause motion, while mechanics combines the principles of kinematics and dynamics to study the motion of objects in more complex situations.

4. What are some key equations in Grade 12 Kinematics?

Some key equations in Grade 12 Kinematics include the equations for displacement (Δx = xf - xi), average velocity (v = Δx/Δt), and acceleration (a = Δv/Δt). Other important equations involve motion graphs, such as the slope of a position-time graph representing velocity and the slope of a velocity-time graph representing acceleration.

5. How can I apply Grade 12 Kinematics in real life?

Grade 12 Kinematics has many practical applications in everyday life. For example, understanding the principles of velocity and acceleration can help you calculate the braking distance of a car or the time it takes for an object to fall from a certain height. It can also be used in sports to analyze the motion of athletes or in engineering to design and test structures and machines.

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