2 Dimensional Uniform motion. A car crossing a road Let us say that there is car crossing a wide...

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2 Dimensional Uniform motion

A car crossing a road

• Let us say that there is car crossing a wide road (15 meters wide) at a speed of 5 m/sec. Finding the time to cross the road should be pretty easy to find.

V = Displacement /Time

Time = Displacement / V

Time = (15 m)/(5 m/sec)

Time = 3sec

5 m/sec

Now Let’s look at a car stalled and floating in a raging river

• In this case the car will not move across the river but move down the river at the same speed as the water.

• How far will the car move down river in 3 seconds if the river flows at a rate of 5 m/sec?

• This also should be fairly easy to do.

V = Displacement/Time

Displacement = V* Time

Displacement = (5 m/sec)* (3 sec)

Displacement = 15 m

5 m/sec0 sec

5 m1 sec

5 m2 sec

5 m3sec

Both happening at the same time• In this case let’s say that a car-boat is trying to cross a 15

meter wide flowing river by pointing directly across the river moving at 5 m/sec, the river also flows at a rate of 5 m/sec. Lets find the time it takes to cross the river.

In the case the car does not move directlyacross the river, or along the river. Instead it moves diagonally across the river.

5 m/sec

However the car is still moving across the river at 5 m/s (just like the car in the 1st example)So it takes 3 seconds to cross.

5 mAnd the car movies down the river just like the second example. So the car travels 15 meters downstream.

Independence of axis• Whenever an object moves along both the X and Y axis the

motion of one axis has nothing to do with the motion of other axis.

• Events that happen along the X axis do not affect events that happen along the Y axis.

5 m/sec

Concept Connection• When working with vectors we often break them up into their

X and Y comments and then combine all the X components together, then combine the Y components together.

We can do this because the Y components do not affect the X components. Because the are along different axis they are independent with each other

Speaking of vectors

• Since velocities are vectors we can find the boat’s resultant velocity by simply adding the velocity of the boat (without the river) to the velocity of the river.

• But this must be done vectorally.V resultant = V boat + V river V boat

V river

V resultant

Applying triangles• We can now use the triangle rules to find the

resultant speed an direction.

(Square root of 50) m/s

(V resultant)2 = (5 m/sec)2 + (5 m/sec)2

(V resultant) = (Square root of 50) m/sec = tan-1(Y/X) = tan-1[(5 m/sec)/(5 m/sec)]

= 45O in the 4th quadrant

= -45O in the 4th quadrant

5 m/sec5 m/sec

Vectors and displacement

• Likewise we can also use the displacements along the X and Y axises to find the resultant displacement

Displacement resultant = Displacement X + Displacement Y

Displacement X

Displacement Y

Displacement

resultant

Appling triangles (again)

(Square root of 450) m/s

(Displacement resultant)2 = (15 m)2 + (15 m)2

(Displacement resultant) = (Square root of 450) m

= tan-1(Y/X) = tan-1[(15 m)/(15 m)]

= 45O in the 4th quadrant

= -45O in the 4th quadrant

3*(Square root of 50) m/s

(Displacement resultant) = 3*(Square root of 50) m

5 m/sec

The power of similar triangles• The displacement triangle and velocity

triangle are similar.• The number of times the displacement

triangle is larger (or smaller) than the velocity triangle is the amount of time it take for the boat to cross the river

• Displacement = Velocity Average*time

• So we can use similar triangles to analyze 2 dimensional systems (with a uniform velocity)

Crossing another river

• Lets say there is a boat that wants to cross a river that is 15 meters wide by pointing directly across and moving at 5 m/sec. The river however flows at a rate of 10 m/sec. How far down stream will the boat land?

5 m/sec

10 m/sec

Distance Downstream

Distance Downstream = 30 m

5m/sec 10m/sec

Distance Downstream

Since the displacement triangle is 3 times larger the time to cross the river is 3 seconds

Distance Downstream

Important Note: Changing the speed of the river did not change the time it took tocross the river, because the two velocities are perpendicular to each other.

5 m/sec

10 m/sec

Distance Downstream

10m/sec

5 m/sec

Cutting across at a diagonal• In this case object moving moves across a field

but at angle and not directly across.

60 m5 m/s

A car crosses a field at a speed of 5 m/s at an angle of 30O as shown. If the car travels 60 meters along the field, find the following…

1) Width of field2) Time to cross the field3) The car’s displacement vector

Vector applications

• In this case since we need to break the car’s velocity vector into it’s X and Y components, so that we can form a velocity triangle.

5 m/s*Cos(30O)

5 m/s*Sin(30O)

2.5 m/s

4.33 m/s

Setting up the triangles• Now that we have the velocity triangle we can now

use similar triangles solve for our displacements

2.5 m/s 4.33 m/s

Width = 103.92 m

2.5 m/s 5 m/s

Displacement

Displacement = 120 m at 30O

Displace

Since the displacement vector is 24 times large than the velocity vector the time is 24 seconds

5 m/s30O

2.5 m/s

4.33 m/s

A more complicated case

• A boat crosses a 120 meter wide river that is flowing at 4 m/sec be heading at 8 m/sec at an angle of 30O into the current. How much time will it take for the boat to cross the river?

8 m/sec

4 m/sec

(8 m/sec)*Cos(30)6.93 m/sec

(8 m/sec)*Sin(30)4 m/sec

4 m/sec

(4 m/sec)*Cos(270)

0 m/sec

(4 m/sec)*Sin(270)

-4 m/sec

6.93 m/sec 0 m/sec

6.93 m/sec(8 m/sec)*Cos(30)

(8 m/sec)*Sin(30)4 m/sec

Resultant velocity = 6.93 m/sec at 0O

First we have to find the resultant velocity of the boat (V resultant = V boat + V river). For this we can use the Vector components (the box method).

(V)*Cos() (V)*Sin()Vectors

V Boat

(8 m/sec) at 30O

V River

(4 m/sec) at 270O

Resultant velocity = 6.93 m/sec at 0O

Resultant displacement = 120 m at 0O

Time = (Displacement/velocity)

Time = (120 m)/(6.93m/sec)

Time =17.32 sec

2 Dimensional Uniform motion. A car crossing a road Let us say that there is car crossing a wide...

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