MOTION VECTORS

==Direction Vectors== When is motion acceleration or deceleration? When the velocity and acceleration vectors align we have acceleration (speeding up). When acceleration vector opposes the velocity vector we have deceleration (slowing down). A negative gradient on a v-t graph does not necessarily mean deceleration. Take the following velocity-time graph and observe the 3 sections: [image:https://i.imgur.com/xCs4Hos.png] We define our space as having a positive and negative direction in one dimension. We start from 0 position and velocity at t=0. '''Section (1)''' – Acceleration in the + direction Displacement = +s (positive) Velocity = +v (positive) Acceleration = +a (positive) '''Section (2)''' – Deceleration in the – direction but velocity still in + direction. The object is still moving in the + direction for this time but it is slowing down. Displacement = +s (positive) Velocity = +v (positive) Acceleration = -a (negative) '''Section (3)''' – Acceleration in the – direction, starting from rest Displacement = -s (negative) Velocity = -v (negative) Acceleration = -a (negative) As you can see, even though the acceleration value is negative, the object is accelerating because the velocity is in the same direction. ==Example: Ball Hitting a Tennis Racquet== [image:http://web.mit.edu/3.082/www/team1_f02/pictures/tennis-ball-rebound-2a.jpg] (Source: http://web.mit.edu/) A tennis ball travelling at 30ms^-1^ to the left hits a tennis raquet and rebounds to the right without changing speed. Determine the change in velocity of the ball. [image:https://i.imgur.com/tns1aO0.png'] The direction of the change in velocity is the same direction as the acceleration and therefore the '''Force''' acting on the ball which produced the change in velocity. ==Example: Bouncing Ball== [image:http://people.rit.edu/andpph/photofile-sci/bouncing-ball-7810.jpg] Let us imagine a bouncing ball being let go at a height h. Assume it is falling without air resistance and once it rebounds, returns to the same height. Also assume the ball is rigid and does not deform on impact. Yes this is a lot of assumptions but here our aim is to imagine a very simple scenario. We can include these things later if we wish and see how they change things. We will initially define our space as downwards being the positive direction and upwards being negative. Since acceleration due to gravity is constant near the surface of the Earth, we expect to see this as a positive value from our graphs The '''distance-time graph''' would continually increase with time as the total length travelled by the object always increases while it is in motion. [image:https://i.imgur.com/OBhUeu6.png] As the ball falls starting at t = 0s, it accelerates, after the ball rebounds, it travels upwards and decelerates and this cycle repeats. The continuous curved line indicates that the speed is constantly changing. [image:https://i.imgur.com/irk9EqD.png] The displacement-time graph tracks the objects actual position in space with time. We will define the ground as our reference position s = 0. At t = 0, the ball starts at height h and the line will only ever go between h and s = 0. We start by accelerating towards the ground, rebounding and then decelerating upwards until we reach h again. [image:https://i.imgur.com/kOqUsqi.png] On the surface of the Earth, the acceleration due to gravity is constant at around g = 9.81ms^-1^. To show this constant acceleration on a speed-time graph is a diagonal straight line. Therefore the size of the gradient of the speed-time graph at all times must be 9.81. At t=0 the ball starts moving from rest as it is dropped and constantly accelerates before hitting the ground. After rebound, travelling upwards it decelerates constantly and comes to rest again at maximum height. [image:https://i.imgur.com/kOqUsqi.png] Velocity is a vector. We have defined our space with downwards as negative and upwards as positive. At t = 0, the ball increases its velocity in the negative direction at a constant acceleration. When it hits the ground at its maximum downwards speed, it quickly changes its direction travelling upwards so the line on the graph jumps the axis. Travelling upwards, it decelerates from maximum speed to zero and then the cycle repeats. Notice that the gradient of the line is constant and has a positive value (from how we defined our space). This is what we expect! The acceleration due to gravity is pointing downwards towards the centre of the Earth at all times so the gradient is always 9.81ms^-2^. Note: The ball takes a little bit of time to deform when hitting the ground to change direction. There would be a large gradient between the lines when the graph changes axes. This large acceleration to change direction is related to the force acting on the ball from the ground.