In solving problems in linear
motion under constant acceleration it is often more advantageous to use
a velocity-time graph than to use the equations of motion.

The use of a velocity-time
graph is beneficial when there are different parts to the motion.

An example might be the motion of a lift in an office block. It could cover the distance from the basement to the roof by accelerating uniformly from rest to it’s maximum velocity, then travelling at constant speed for the middle section before decelerating, at a constant rate, to rest at the roof garden.

You might like to try a
graph like this in your graph copy.

Here is the data required,
in the form of a little story.

A lift starts from rest
in the basement of a hotel and accelerates uniformly for 15 seconds to
reach its maximum speed in 25 seconds. The lift then travels for 30 seconds
at this maximum speed for before decelerating uniformly to rest in a time
of 10 seconds.

Tan A =
= f

So, the acceleration is the slope of the graph.

Area of rectangle 1 = ut

Area of triangle 2 = ½(v-u).t
= ½(ft).t = ½ft^{2}

Area 1 + area 2 = ut +
½ft^{2} = s

The area under the graph
gives the distance travelled.

Two very important points:
**Acceleration = Slope
of line**
**Distance = Area under
graph**

Another way of looking at
this is:

The equation of a line
in slope-intercept form is *y* = *mx* + *c*

The equation for final
velocity is v = u + ft

If we rewrite this as v
= ft + u and compare v = ft + u with *y* = *mx* + *c* we
can see by inspection that the slope m = f i.e. the slope gives the acceleration.

The area of the graph can
be calculated using the area of a trapezium.

Area ==
s

i.e. Distance = area under
the graph.