Vector and Scalar Quantities (recap on National 5 Physics)
A scalar quantity is fully described by its magnitude (size) and unit, e.g.
unit
time = 220 s
quantity
magnitude
A vector quantity is fully described by its magnitude, unit, and direction, e.g.
unit
Force = 800 N upwards
quantity
direction
magnitude
Distance and Displacement
Distance is a scalar quantity. It measures the total distance travelled, no matter in which direction.
Displacement is a vector quantity. It is the length measured from the starting point to the finishing
point in a straight line. Its direction must be stated.
Example
A girl walks 3 km due north then turns and walks 4 km due east as shown in the diagram.
000˚
4 km
FINISH
090˚
270˚
3 km
START
180˚
Calculate
(a) The total distance travelled
(b) The girl’s final displacement relative to her starting position.
Solution
(a) Total distance = 3 km + 4 km
= 7 km
(b) (to calculate displacement, we need to draw a vector diagram)
(this solution involves using Phythagoras and trig functions (SOH-CAH=TOA), but you can
also solve these types of problems using a scale diagram).
4 km
FINISH
Magnitude of displacement =
= 5 km
3 km
θ
Resultant
displacement
Θ= 53˚
START
Resultant displacement = 5 km at 053˚
Speed and Velocity

Speed is a scalar quantity. As discussed above, speed is the distance travelled per unit time.
(

Velocity is a vector quantity (the vector equivalent of speed). Velocity is defined as the
displacement per unit time.
(


)
)
Since velocity is a vector, you must state its direction.
The direction of velocity will be the same as the displacement.
Example
The girl’s walk in the previous example took 2.5 hours.
Calculate
(a) The average speed
(b) The average velocity for the walk, both in km-1
Solution
(a) Distance = 7 km
Time = 2.5 hours
(
(b) Displacement = 5 km (053˚)
)
(
)
Time = 2.5 hours
at 053˚
(When performing calculations on speed and velocity like the above, it is best to write the equations in
words as shown, to avoid confusion with symbols. This communicates your understanding in the clearest
way).
Rectangular Components of a Vector

Any vector v can be split up (resolved) into a horizontal component vh and vertical
component vv.
is equivalent to
Example
A shell is fired from a cannon as shown.
Calculate
a) the horizontal component of velocity
b) the vertical component of velocity.
a) vh = v cos u = 50 cos 60° = 25 m s-1
b) vv = v sin u = 50 sin 60° = 43.3 m s-1
So
is equivalent to
Acceleration


Acceleration is defined as the change in velocity per unit time.
The unit is metre per second squared, m s-2.
a =
v - u
t
where
v = final velocity
u = initial velocity
t = time taken
Measuring acceleration
Acceleration is measured by determining the initial velocity, final velocity and time taken. A double
mask which interrupts a light gate can provide the data to a microcomputer and give a direct reading
of acceleration.
Graphs of motion



Displacement-time
Velocity-time
Acceleration-time
Interpreting displacement–time graphs

The gradient of a displacement–time graph gives the velocity of the object.
Interpreting velocity–time graphs


The gradient of a velocity–time graph gives the acceleration of the object:
the steeper the line the greater the acceleration of the object.
The area under the graph is the displacement.
Interpreting acceleration-time graphs

The area under the graph gives us the velocity of the object
The following graphs show the relationship between the three graphs
Stationary object
Constant velocity
Acceleration-Time graph
Constant acceleration
Acceleration-Time graph
Example
The graph below represents an object moving with a positive velocity of 5 m s–1, which is
accelerating at a constant rate. After 300 s the object is moving with velocity of 35 m s–1. A constant
acceleration means the velocity is increasing at a constant rate.
The acceleration of the object can be found from the gradient of the line.
The displacement of the object can be determined by calculating the area under the graph.
2
1
displacement = area under graph
= area 1 + area 2
= (300 x 5) + ( ½ x 30 x 300)
= 1500 + 4500
= 6000 m
Example
Sketch of velocity-time graph for a ball thrown upwards
Equations of motion
where:
v = u + at
s = ut +
2
u - initial velocity of object at time t = 0
v - final velocity of object at time t
a - acceleration of object
1 2
at
2
t - time to accelerate from u to v
2
v = u + 2as
s - displacement in time t.
These equations of motion apply providing:


the motion is in a straight line
the acceleration is constant.
When using the equations of motion, note:



the quantities u, v, s and a are all vector quantities
a positive direction must be chosen and quantities in the reverse direction must be given a
negative sign
a deceleration will be negative, for movement in the positive direction.
Derivation of equations of motion
1. v=u + at
The velocity - time graph for an object accelerating uniformly from u to v in time t is shown below.
a =
v -u
t
Changing the subject of the formula gives:
v = u + at
--------- [1]
2. s=ut + ½at2
Consider the velocity-time graph of an object accelerating from initial velocity u to final velocity v in
a time t.
( )
(
)
(But from equation 1, v = u+at)
velocity / ms
-1
The displacement, s, in time t is equal to the area under the velocity time graph.
v
B
u
A
0
0
----------- [2]
3. v2 = u2 + 2as
(square both sides)
(
)
(
)(
)
(multiply out brackets)
(take common factor 2a from 2nd and 3rd term)
(
)
------------ [3]
(and from equation[2], s=ut + ½at2)
t
time / s
Objects Launched Upwards



At the instant an object is launched upwards, it is travelling at maximum velocity. As soon as
the object starts to travel upwards, gravity will accelerate it towards the ground at -9.8 ms-2.
As a result, the upward velocity of the object will eventually become 0 ms-1.
This happens at its maximum height.
Example
A spring-powered toy frog is launched vertically upwards from the ground at 4.9 ms-1.
(a) What will be the velocity of the toy frog at its maximum height?
(b) Calculate:
(i) the time taken for the toy frog to reach its maximum height;
(ii) the maximum height.
Solution
s=?
u = 4.9 ms-1
v=?
a = -9.8 ms-2
t=?
a)At maximum height v = 0 ms-1.
b)(i)
v = u + at
0 = 4.9 + (-9.8t)
0 = 4.9 - 9.8t
9.8t = 4.9
t = 4.9/9.8
t = 0.5 s
(ii)
s = ut + ½ at2
s = (4.9 x 0.5) + (0.5 x -9.8 x 0.52)
s = 2.45 + (-1.225)
s = 1.2 m i.e. 1.2 m upwards, so height = 1.2 m