# VELOCITY-TIME GRAPHS

#### branchMECHANICS (MOTION)

Coming from ACCELERATION
=Velocity-Time Graphs= [image:http://i.imgur.com/OsHIGch.png?1] On a velocity-time graph a '''horizontal (flat) line''' indicates the object is travelling at a '''constant speed'''. A '''straight diagonal line''' indicates the object's velocity is changing. In the graph on the left, the line sloping upwards shows the object is '''accelerating''' and the line sloping downwards in this case towards v = 0, shows it is '''decelerating'''. The negative value of the gradient gives the negative value for the acceleration (or deceleration), BUT, a negative value does not always mean slowing down! See the next section on ''''DIRECTION VECTORS'''' for elaboration. Be careful here. The general rule of thumb is if the object is going from a high speed to a low speed, it is decelerating. ==Calculating Acceleration from a Velocity-Time Graph== The average acceleration can be calculated for any part of a journey by taking the '''change in velocity''' and dividing by the '''change in time''' for that part of the journey. [image:http://i.imgur.com/GIXpoaO.png] ==Calculating the Distance Travelled from a Velocity-Time Graph== [image:http://i.imgur.com/Vcxcoon.png] The total distance travelled by an object can be determined by calculating the area underneath the velocity time graph. Start by dividing the graph into sections that consist of simple triangles and rectangles as shown below in blue and red. Calculate the area of each shape as shown below. [image:http://i.imgur.com/49qRk1v.png?2] You can then simply add the areas together and the total area represents the total distance travelled. E.g. the graph above has a total area of: $(1/2 \times 10 \times 20) + (20 \times 20) + (1/2 \times 40 \times 20) = 900$ Therefore the total distance travelled was 900m. ==Describing Velocity-Time Graphs== When describing the motion of an object try to be as detailed as possible. For instance... [image:http://i.imgur.com/I3wusPk.png] During Part A of the journey the objects velocity increases by 8m/s in 4s. It is '''accelerating''' at a rate of 2ms^-2^ During Part B of the journey the object travels at a '''constant speed''' of 8ms^-1^ for 3s During Part C of the journey the objects velocity decreases by 8m/s in 3s. It is '''decelerating''' at a rate of -2.7ms^-2^ === === Note that students often confuse '''distance-time''' graphs with '''velocity-time''' graphs. In an exam, make sure you know which one you are dealing with.
Credit: Tristan O'Hanlon, Ben Himme