1.7 determine acceleration from the gradient of a velocity-time graph

I find it helps to remember 'rise over run', all this means is that in order to work out the acceleration all you need to do is do velocity / time.

In the example above...

The acceleration represented by the sloping line is 2m/s². This is because the object increases its velocity from 0m/s to 8m/s in 4s therefore its acceleration is 8 ÷ 4 = 2m/s^2.

1.8 determine the distance travelled from the area between a velocity-time graph and the time axis

The area under the line in a velocity-time graph represents the distance travelled. To find the distance travelled in the graph above, we need to find the area of the light-blue triangle and the dark-blue rectangle.

The area of light-blue triangle; the width of the triangle is 4 seconds and the height is 8 metres per second. To find the area, you use the equation:
area of triangle = ¹⁄₂ × base × height
so the area of the light-blue triangle is ¹⁄₂ × 8 × 4 = 16m. 

Area of dark-blue rectangle; the width of the rectangle is 6 seconds and the height is 8 metres per second.
 So the area is 8 × 6 = 48m.

Area under the whole graph; the area of the light-blue triangle plus the area of the dark-blue rectangle is:
16 + 48 = 64m covered by the object.

1.6 plot and interpret velocity-time graphs

Interpreting the graph...

straight line = constant speed

straight line going upwards = steady, constant, increase in velocity (constant acceleration)

straight line going downward = steady, constant, decrease in velocity (constant deceleration)

curved line = changing acceleration

straight line at base of graph = stationary (no velocity)
The steeper the slope, the faster the acceleration/deceleration

In the example above...

The object is stationary for the first 3 seconds, then has a steadily increasing speed for 2 seconds. For the next 3 seconds it has a constant speed, and for the last 2 seconds it has a steadily decreasing speed.

1.5 know and use the relationship between acceleration, velocity and time

acceleration = change in velocity/time taken 

1.4 describe experiments to investigate the motion of everyday objects such as toy cars or tennis balls

toy car:

 record the time it takes for a toy car to travel a measured distance, plot on a distance-time graph. (e.g. the toy car takes 10 seconds to travel 4 metres). repeat at different speeds and plot graphs. compare results.

tennis ball:
record the time it takes a tennis ball to be thrown 10m, repeat 10 throws from the same person (to make it a fair test, as some people will throw faster/slower). Plot results on a graph.

1.3 know and use the relationship between average speed, distance moved and time:

average speed = distance/time

1.2 plot and interpret distance-time graphs

The vertical axis of a distance-time graph is the distance travelled from the start, and the horizontal axis is the time taken from the start.

How to interpret...

When an object is stationary, the line on the graph is horizontal. 

When an object is moving at a steady speed, the line on the graph is straight (but not horizontal)

The steeper the line the faster the object is travelling (as it is covering more distance in a shorter time), and vice-versa if the line is not steep, the object will be traveling slower.

If the line is sloping downwards, the object us returning to the start.

In this example: The blue line is steeper than the red line because it represents an object moving faster than the object represented by the red line. The red lines on the graph represent a journey where an object returns to the start again.