Math Is Fun Forum

paulb203

Member

Registered: 2023-02-24

Posts: 321

Reference Frames

I’m stationary on a railway platform, WRT (with reference to) the others sitting or standing on the platform.
A train is travelling on the track in front of me at 30m/s
A skateboarder on the platform opposite is travelling, WRT me, at 5m/s, in the same direction of the train

What’s the speed of the boarder WRT the train?

What’s the speed of the train WRT the boarder?

What’s the speed of the boarder WRT to me?

What’s the speed of the train WRT to me?

*

A skateboarder on the train is travelling at 5m/s in the direction of the train’s travel.

What is their speed WRT to me?

What is their speed WRT to the boarder on the platform?

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Prioritise. Persevere. No pain, no gain.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Reference Frames

The way I was taught to do this is to bring the 'WRT-object' to  rest by applying an equal and opposite speed to it, and apply the same speed to everything else.  As you are stationary that means the WRTs in the parts 3 and 4 are just the given information with no change needed.

For part 1 bring the train to rest by applying a speed of 30 m/s in the opposite direction.  Let's say, to avoid confusion, that the train is heading North, and so is the boarder.  So make the whole world go South at 30 m/s so that the train is brought to rest. The boarder is also given a speed South of 30 m/s so its speed North is now (5 - 30) m/s  ie. 25 m/s South.  If a person is on the train they think they're stationary and the whole world is travelling South at 30 m/s. Except the boarder of course who seems to be travelling South at 25 m/s.

You can try this yourself for part 2. You should get the exact opposite answer to part 1

A skateboarder on the train is travelling at 5m/s in the direction of the train’s travel.

So add the boarder's speed to the train's to get 30 + 5 m/s

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

paulb203

Member

Registered: 2023-02-24

Posts: 321

Re: Reference Frames

Thanks, Bob.

So,
Q What’s the speed of the boarder WRT the train?
A Bring train to rest (apply -30m/s).
Same to boarder (5m/s-30m/s)
Boarder; -25m/s WRT train

Q What’s the speed of the train WRT the boarder?
A Bring boarder to rest (apply -5m/s)
Same to train (30m/s-5m/s)
Train; 25m/s WRT boarder

Q What’s the speed of the boarder WRT to me?
A No need to bring me to rest as I’m already at rest. Don’t apply anything
Boarder 5m/s WRT me

Q What’s the speed of the train WRT to me?
A As above
Train; 30m/s WRT me

*

The idea of moving at a negative speed is a curious one to a physics novice like me. Is this a helpful way of beginning to grasp it;

The train is travelling at 30m/s
The boarder, WRT train, is travelling at -25m/s
Assuming they start at the same point, after 1 second the boarder will be 25m behind the train
After 2 seconds the boarder will be 50m behind
Which seems to fit with him travelling at -25m/s (WRT train)

D=ST
D=-25m/s(2s)
D=-50m

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Prioritise. Persevere. No pain, no gain.

paulb203

Member

Registered: 2023-02-24

Posts: 321

Re: Reference Frames

P.S.

Is the object following the phrase, "with respect to", always the reference frame?

E.g, What's the speed of the boarder with respect to the train. Is the train the reference frame?

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Prioritise. Persevere. No pain, no gain.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Reference Frames

Your calc leading to -50 is a good one. 

Yes to the frames question.

Speed is a scalar quantity so I wouldn't expect it to be negative. But velocity is a vector so direction is needed. When you start such a question decide which way is positive and if any calc leads to a negative it means it's going the other way.

eg. A ball is thrown vertically upwards at 27 m/s. Taking g as 10, what is its height after 5 seconds.

I'll take 'up' as positive.

g = -10; t = 5; u = 27

using s = ut + 0.5at^2

height = 27 x 5 - 0.5 x 10 x 5^2 = 135 - 125 = 10m (above the ground)

I'll take 'down' as positive

g = 10; t = 5; u = -27

height = -27 x 5 + 0.5 x 10 x 25 = -135 + 125 = -10m  (As down is positive this ball must be 10 m above the ground.

Suppose u = 7, and I throw the ball up leaning sightly over a cliff.

Take 'up' as positive.

height = 7 x 5 - 0.5 x 10 x 25 = 35 - 125 = -90m   The ball has gone up, come down and carried on downwards below my level down the cliff.

When was the ball again level with my position?

t = ?; u = 7; g = -10; s = 0

0 = 7t - 0.5 x 10 x t^2    =>       7t - 5t^2 = 0       t(7-5t) = 0 So we have two solutions, t = 0 and t = 7/5

We should expect the first as we know the ball was at that height when we threw it and started timing. The second is the one that answers the question.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob