Concepts Quantities relating to motion are often vectors, but scalars can be used in the case of straight-line motion, or if the direction is not important in a particular situation. Sometimes different names are used to distinguish them. Scalar Vector How far? distance displacement m How fast? speed velocity How is the speed changing? acceleration acceleration ms-2 Unit ms-1 Instantaneous and average values The formula speed = distance / time is a simplification. It is important to distinguish between an average and an instantaneous value. Average speed (velocity) Average speed = total distance total time Instantaneous speed (velocity) Instantaneous speed = rate at which distance is changing with time. This can be found from the formula speed = distance / time, for a time which tends towards zero, or from the gradient of a distance-time graph at a point in time. If the graph is a curve, than the gradient of the tangent at the point in time gives the value. The same principle applies to velocity (= displacement / time) and acceleration ( = change in velocity / time) Motion Graphs velocity ms-1 10 5 2 4 6 8 10 time / s Sketch a-t and d-t graphs for this motion. Interpreting Motion Graphs (gradient) Section 1 of the Graph The velocity increases from 0 to 10 ms-1 in a time of 2s. We can calculate the average acceleration, but the fact that this section is a straight line shows that the acceleration is constant. a = (10 - 0) / 2 = 5 ms-2 This value applies at every point from 0s to 2s. Section 2 of the Graph In this section the velocity is not changing. a = 0 for the whole time. Section 3 of the Graph A constant negative gradient shows a constant negative acceleration. a = (0 - 10) / 3 = -3.3 ms-2 Notice that the velocity is still positive. Section 4 of the Graph Again, the velocity is not changing. Its value of 0 is still a constant. Again a = 0. The gradient of the v-t graph gives the acceleration; the gradient of the d-t graph gives the velocity. Interpreting motion graphs (area) Section 1 of the Graph The average speed during the first 2s is 5 ms-1. So the distance covered is 5 ms-1 x 2s = 10m. This corresponds to the area of the shape under the graph line. Section 2 of the Graph Here the speed is a constant 10 ms-1 for a time of 5s. The distance covered is 50m. Note that this is an additional 50m to the 10m covered in the first section. When deriving a d-t graph from a v-t graph, make sure to add the values on to the previous total. In general, the area of a trapezium under the graph corresponds to the formula for average speed. A triangle or rectangle are special cases. velocity v u s = (u + v) t 2 Δt time (u+v)/2 = average speed and also the average of the parallel sides Δt = distance between the parallel sides and also the time. The area under a section of a v-t graph represents the additional distance covered during the time. The area under a section of an a-t graph represents the change in velocity during that time. Note that areas below the time axis are negative and represent negative displacements or velocity changes. In summary: Uniform Acceleration There are many situations in which an object is subject to uniform acceleration. i.e. an acceleration which is constant in magnitude and direction. This is the case when the body is subject to a constant force, such as gravity. A series of equations, based on the work of Newton, relate the various kinematic quantities. These are commonly referred to by students as the "suvat" equations, from the common symbols for the variables involved: s = displacement (m) u = initial velocity (ms-1) v = final velocity (ms-1) a = acceleration (ms-2) t = time (s) A v-t graph for a body moving with uniform acceleration looks like this: The "suvat" Equations From the gradient: a = (v-u) / t or v = u + at From the area under the graph (Area of a trapezium): s = (u+v) t 2 Newton's equations for uniformly accelerated motion. From the area under the graph (Rectangle plus Triangle): s = ut + ½(v-u) t but v-u = at so s = ut + ½at2 Squaring the first equation: v2 = (u+at)2 = u2 + 2uat + a2t2 = u2 + 2a(ut + ½ at2) so v2 = u2 + 2as Free Fall and Air Resistance Acceleration of Free Fall The force of gravity on a body near the Earth's surface causes an acceleration of approx. 9.8 ms-1 (providing air resistance is negligible). This is given the symbol g as it also describes the gravitational field strength. (See Topic 6) Air resistance When a body moves through a fluid (liquid or gas) fluid friction acts in the direction opposite to the velocity of the body. In the common case of a solid body moving through the air, this is called "air resistance." The magnitude of this force depends on the shape of the body and its velocity. If a body is falling, the downward force of gravity is constant, but the upward force of air resistance increases as the velocity increases. Eventually a velocity will be reached where the air resistance equals the force of gravity. With a net force of zero, there is no net acceleration. The velocity at which this occurs is called the "terminal velocity." Relative Velocity If a body is moving with a constant velocity, this velocity appears different to a stationary observer and one who is also moving. In all cases (except close to the speed of light) the relative velocity of the observed body is equal to its absolute velocity (as observed from a stationary point) minus the absolute velocity of the observer. vr = v - vo Relative velocity in 1-D Since the only directions are forwards and backwards, it is usual to use + for forwards and - for backwards. Algebraic calculations will give relative velocity. e.g. You are driving at 20 ms-1 along a straight road. A car approaches from the opposite direction with a speed of 15 ms-1 (so -15 ms-1). The speed of the approaching vehicle, relative to you is: velocity of the other car - your velocity = (-15) - 20 = -35 ms-1 i.e. 35 ms-1 in the direction you are coming from Relative velocity in 2-D In this case vector subtraction must be used. It is often useful to use the notation vAB for the velocity of object A relative to B. The general vector sum vAB = vAF + vFB, where F is a fixed point of reference can be used. (vFB = -vBF) Example Car 1 is travelling north at 17 ms-1 when car 2 crosses an intersection ahead at 10 ms-1 from W to E without giving way. Find the velocity of car 2 relative to car 1.