Vectors for Mechanics

j

i

e.g. A velocity v is given by

v = (3 + 4 ) m s -

1

i jx

y

j

i3

4v

Instead of drawing diagrams to show vectors we can use unit

vectors. A unit vector has magnitude 1.

The unit vectors and are

parallel to the x- and y-axes respectively.

i j

In text-books single letters for vectors are printed in bold but we must underline them.

v = 3 + 4 i j

The magnitude of velocity is speed, so, using Pythagoras’ theorem,

v = 32 + 42

No or in magnitude

i jv = 32 + 42

So, if we have the unit vector form, we use the numbers in front of and

i j

x

y

3

4v

j

i

Tip: Squares of real numbers are always positive so we never need any minus signs.

v = 5

We can write v or v for

speed.

v = + i jThe direction of the vector is found by using trig.

tan q = q = 53·1 ( 3

s.f. )

BUT beware !

3 443

x

y

3

4v

j

i

If we need the direction of a vector when unit vectors are used, we must sketch the vector to show the angle we

have found.

v = -3 - 4 i jSuppose

Without a diagram we get tan q = -3

-4

q = 53·1 ( 3 s.f. )

So again

But, the vectors are not the same !

v = + i j q = 53·1 ( 3 s.f. )

3 443For we have

3

4

q

i j3 443v = +

3

4v = -3 - 4 i j

q

Equations of Motion for Constant Acceleration

We can use a velocity-time graph to find some equations that hold for a body moving in a straight line with constant acceleration.

velocity (ms-1)

time (s)

u

v

t0

Suppose when the time is 0 . . .

At any time, t, we let the

velocity be v.

the velocity is u.

Ans: The gradient gives the acceleration.

Remind your partner how to find acceleration from a velocity-time graph.

Constant acceleration means the graph is a straight line.

We can use a velocity-time graph to find some equations that hold for a body moving in a straight line with constant acceleration.

velocity (ms-1)

time (s)

u

v

t

a = v - u t

So,

v - u

0

Suppose when the time is 0 . . .

the velocity is u.

t

From this equation we can find the value of any of the 4

quantities if we know the other 3.

At any time, t, we let the

velocity be v.

Constant acceleration means the graph is a straight line.

a = v - u t

a t = v - u

v = u + a

t

Multiplying by t: u + a t = v

We usually learn the formula with v as the “subject”.

The velocity, u, at the start of the time is often called the initial velocity.

Displacement and Velocity using Unit Vectors

e.g. A ship is at a point A given by the position vector

r A = (-4 + 3 ) kmi j

Solution:

Find

(a) the displacement of B from A, and

After half-an-hour the ship is at a point B.

We can solve this problem without a diagram, but a diagram can help us to see the method.

iThe ship has a constant velocity of 6 km h -1.

(b) the position vector of B.

O

y

x4

3

A x6 i

After half-an-hour the ship is at a point B.

B xdisplacement = velocity time

= 3 km

i

s = 6 i

(-4 + 3 ) i j + 3 i=

(- + 3 ) kmi j

r A = (-4 + 3 )

km

i jA: Velocity v = 6 km h

-1i

r B = r A + s

r B =

r Br A

Solution:

(a) Find the displacement of B from A.

0·5(b) Find the position vector of

B.

Constant velocity s