Vectors and relative motion

Vectors By now, you should know some simple vectors: displacement, velocity, force and acceleration. All of these quantities have magnitude (size) and direction. We represent vectors in different ways, but one of the most common is to use an arrow. The size of the arrow represents magnitude, and the direction shows the direction of the vector This vector shows velocity at 2ms -1 to the right

Parts of a system It’s important, when talking about relative motion, to consider the parts of the system that we’re interested in. For example, a person throws a ball straight to someone else on a windy day. We all know that the ball will go wide – but to talk about it using physics, we have to define the system about which we are talking In this case, we have the ball, the ground and the air – they are the three parts of the system, with us as an observer

Throwing a ball on a windy day This diagram shows a ball thrown on a windy day, as seen from above: Us, observing As you can see, the ball and the air are moving. The overall velocity of the ball is the velocity relative to the ground. That is, when we stand still and watch from the ground, it’s what we see Velocity of the ball as thrown Velocity of the wind Overall velocity of the ball we observe

Common sense and physics We all know, in the previous example, that if we throw a ball straight to someone on a windy day, that it probably won’t go straight to them. The trick of physics is to find out both why, so that we can explain it, but also by how much, so that we can find solutions using mathematics We could use our common sense to find a practical solution, and, indeed, most people can adjust the direction of a throw to take account of any wind present. But we need to use mathematics to find out exactly.

Rowing across a river Let’s say I am in a row boat about to cross a 25m wide river. I am not a good rower, so I start pointing straight at my friend across the water and row at a steady speed of 4.5ms -1 the whole way. However, a steady current of 1.2ms -1 is also flowing, at right angles to me. Here is a diagram: 4.5ms ms -1 Common sense already tells me I will end up downriver from my friend – but how far? Usually, we’ll be asked to find the actual velocity first

Mathematical solution – scale diagram There are two ways to find a solution. First, I will use a scale diagram 1.Set a scale. In this case, I will set 1ms -1 :2cm 2.Draw the vectors I know, scaling them appropriately. In this case, I will draw the velocity of my boat relative to the water as an arrow 4.5x2=9cm long, and the velocity of the water relative to the bank as an arrow 1.2x2=2.4cm long 3.Draw the resultant vector 4.Measure its length 5.Reverse the scale

Scale diagram continued When I measure the length of this side, I find it is 9.3cm long. When I measure the angle, with a protractor, I find it to be 15 o relative to the water When I reverse the scale, I find that the velocity vector must be 4.7ms -1

Mathematical solution – Pythagoras and trigonometry To solve this problem, I need to use simple mathematics. 1.Draw a diagram, not necessarily to scale 2.Label the vectors and identify the opposite, adjacent and hypotenuse sides 3.Use Pythagoras to find the length of the side you are looking for 4.Now use trigonometry to find the angle, by using sin, cos or tan

Pythagoras and trigonometry continued Adjacent = 4.5ms -1 HypotenuseOpposite 1.2ms -1 Using the formula for pythagoras: Hyp = √(Adj 2 + Opp 2 ) = √( ) = 4.7ms -1 tan θ = opp/adj θ = tan -1 (1.2/4.5) = 15 o

Finding where you end up To find your landing point, you need to work out the time taken to cross the river, using the velocity relative to the bank. v = d/tt = d/v = 25/4.5 = 5.55s Now find the distance the river current will take you in that time: v = d/td = vt = 1.2 x 5.55 = 6.7m