Openphysics/Motion

Relative Velocity Video (17 mins YouTube) - recorded in a NZ year 12 physics class during 2011. Interactive board using OneNote for writing the notes and CamStudio to record the video.

Discuss why a car can go faster around a banked corner compared to a non-banked corner.

Because of the added centripetal force due to the horizontal component of reaction force. The car is able to go faster around a banked corner. Where as when a car goes around a normal corner the reactant force is upward instead of horizontal, the car is not able to go as fast around the non-banked corner.

Motion
The motion of objects can be described and measured using various quantities such as distance, speed and acceleration. Motion also involves the direction in which the object is moving. When the direction is important, the scalar quantities of speed and distance must be replaced by the vector quantities of displacement and velocity.

Displacement, velocity and time
Distance and displacement both have the symbol ‘d’, and both measure changes in position; distance however, is a scalar quantity, and only involves the size of the movement – not the direction – whereas displacement involves both the size, and the direction of motion from a reference or starting point. Similarly, speed only measures the size of motion (change in distance-over-time) whereas velocity measures the size and direction of motion (change in displacement-over-time.)

The formula’s are as follows: Using vector quantities: d = displacement v = velocity t = time Using scalar quantities: d = distance v = speed t = time

Instantaneous and Average speed
Instantaneous speed is the speed something has at any one instance “the dog is running at 2 metres per second”. Objects don’t always move at steady constant speeds, “Ted’s average speed in the marathon was 6 kilometres per hour”. So the average speed is found by dividing the total distance (or displacement) by the time interval - average speed = total distance/time interval Most often, it is the average speed that is required for speed/distance/time and velocity/displacement/time problems

Example1:
A car travels 100 metres north in 20 seconds. What is the car’s average velocity? Displacement = 100 m velocity =? time = 20 s v = d/t v = 100/20 v = 5 Displacement and time are measured in this case in metres and seconds, so velocity is measured in metres-per-second, or m/s, or ms-1, therefore, the velocity of the car = 5ms-1 north.

Example2:
A boat travels south downriver at a constant speed of 30kmh-1, for 180 minutes. Find how far the boat travelled (its displacement) in kilometres. Displacement =? velocity = 30kmh-1 time = 180 minutes The time is given in minutes, but the velocity is kilometres-per-hour, so the time quantity must be converted into hours; 180/60 = 3 hours. d = v*t d = 30*3 d = 90 Velocity and time are measured in kilometres-per-hour and hours, so displacement is measured in kilometres – as asked - so the boats displacement = 90km south.

Example3:
A parachutist reaches terminal velocity of 100 kmh-1 and then continues to fall 700 metres before pulling his parachute. How long in seconds is the parachutist falling for? Velocity is given in kilometres, and displacement in metres; 700 metres = 0.7 kilometres. t= d/v t=0.7/100 t = 0.007 (hours) 0.007*60 = 0.42 minutes 0.42*60 = 25.2 seconds The parachutist falls at terminal velocity for 25.2 seconds before activating his parachute

= Acceleration =

Acceleration is both a scalar and vector quantity. When an object changes in velocity (whether it be slowing down, speeding up or changing direction) the object is said to be accelerating. Acceleration, ‘a,’ is found by the ‘change in velocity’ divided by the ‘time taken for the change in velocity’. The term acceleration applies to both increases and decreases in velocity (when it is decreasing, then there is deceleration.) When we travelling in curved path, or one that changes direction, then there is also acceleration –even if we’re moving at a constant speed, because our velocity (directional speed) is changing. It is for this reason that we call acceleration the ‘change-of-velocity-over-time’, as this encompasses changes in speed, and in direction. If an object travels at a constant velocity – there is no acceleration – the velocity isn’t changing at all. The object has balanced forces acting on it, so with regards to acceleration, it may as well be stationary.

a = v/t    v = a*t      t = v/a
Acceleration can also be found when the force acting on an object and its mass are known*. Fnet = m*a a = Fnet/m m = Fnet/a *information on forces is available below The units of acceleration vary depending on that of time and velocity: a = 10kmh/5h a = 2 kilometres per hour per hour, or 2km h or, (most commonly) 2kmh-2 If the units are different, it would look like this: a = 10kmh / 10s a = 1 kilometre per hour per second, or 1km/h /s, or 1km-1 s-1

Example1:
A bird flying southeast accelerates from 20kmh-1 to 25kmh-1 in 30 minutes. What is the bird’s acceleration? Change in velocity = 25-20 = 5 time = 30 minutes, = 0.5 hours a = v/t a = 5/0.5 a = 10 kmh-2

Example2:
A truck is towing a 1,000 kilogram car with a force of 20,000 Newton’s. What is the car’s acceleration? Fnet = m*a     a = Fnet/m      a = 20,000 / 1,000     a = 20ms-2 Deceleration problems

Example3:
A ball is kicked east by a football player. The moment the boot makes contact with the ball it travels at 20ms-1, the ball then immediately decelerates, and comes to a halt 10 seconds later. I) What is the ball’s deceleration? II) How far does it travel before coming to a stop?

I) vi = 20ms-1, vf= 0 t = 10 s Change in velocity = 0 - 20 = -20 ms-1 a = v / t a = -20 / 10 a = -2ms-2 deceleration = 2ms-2 Note: the ball immediately decelerates because as soon as it leaves contact with the boot, there are unbalanced forces acting on it. There are no forces pushing it forwards anymore, but there are still forces acting to slow it down, and bring it to a stop (gravity and air resistance while it is in the air, and friction when it rolls on the ground) therefore, the ball immediately decelerates. II) Average velocity = (20-0) / 2 = 10ms-1 d =v*t d = 10*10 d = 100 metres The ball’s displacement = 100 metres east.

Further exercises on motion *no answers* are available here http://hrsbstaff.ednet.ns.ca/nmaclell/Science%2010%20and%20O2/pdf%20physics%20science%2010%20part%203.pdf