TJC Unit 3 Motion & Forces

1 i Unit 3: Motion & Forces Learning Outcomes Students should be able to: (a) show an understanding of and use the terms position, distance, displacement, speed, velocity and acceleration. (b) use graphical methods to represent distance, displacement, speed, velocity and acceleration. (c) identify and use the physical quantities from the gradients of position-time or displacement-time graphs, and areas under and gradients of velocity-time graphs, including cases of non-uniform acceleration. (d) derive, from the definitions of velocity and acceleration, equations which represent uniformly accelerated motion in a straight line. (e) solve problems using equations which represent uniformly accelerated motion in a straight line, e.g. for bodies falling vertically without air resistance in a uniform gravitational field. (f) show an understanding that mass is the property of a body which resists change in motion (inertia). (g) define and use linear momentum as the product of mass and velocity. (h) state and apply each of Newton’s laws of motion: 1st law: a body at rest will stay at rest, and a body in motion will continue to move at constant velocity, unless acted on by a resultant external force; 2nd law: the rate of change of momentum of a body is (directly) proportional to the resultant force acting on the body and is in the same direction as the resultant force; and 3rd law: the force exerted by one body on a second body is equal in magnitude and opposite in direction to the force simultaneously exerted by the second body on the first body. (i) recall the relationship that resultant force 𝐹 = 𝑚𝑎 for a body of constant mass, and use this to solve problems. Name : ________________________________ Class : _________

2 1 Introduction Kinematics is the study of objects in motion. Motion occurs all around us. We see it in the everyday activity of people, of cars on the expressway, of javelins thrown by sportsmen, and even in the motion of comets. In Kinematics, we describe motion in terms of space and time while ignoring the agents that caused that motion. In the later part of this unit, we study the effects of forces on the motion of bodies. 2 Definition of quantities LO(a) 2.1 Position, Distance and Displacement Position is defined as the location of the object at any given time. For one-dimensional motion, we often choose the x axis as the line along which the motion takes place. Then the position of the object at any moment is given by its x -coordinate. If the motion is vertical, as for a dropped object, we usually use the y axis. Consider a particle, P, moving from position A to position B. • Dotted lines represent the path taken by the particle P. • The distance travelled by P between points A and B is the length of the dotted path. • The displacement from point A is represented by the vector, s. • The magnitude of the displacement of P is the length of the displacement vector. Distance, x Displacement, s Definition Distance is the total length of the path taken. Displacement is defined as the change in position of the object. It is the distance travelled in a straight line in a specified direction from a given reference point. Type of quantity scalar vector SI unit metre (m) metre (m) s P Starting position, A Ending position, B

3 2.2 Speed and velocity (v) Speed v Velocity v Definition Speed is the rate of change of distance. Velocity is the rate of change of displacement. Type scalar vector SI unit m s-1 m s-1 Speed: * Instantaneous speed v is the rate of change of distance at that instant. dsv dt= * Average speed total distance total time sv t ==  * A body that travels equal distances in equal time intervals is said to be moving with constant or uniform speed. Velocity: * Instantaneous velocity is the rate of change of displacement at that instant. dsv dt= * Average velocity total displacement total time sv t ==  * A body that travels equal displacement in equal time intervals is said to be moving with constant or uniform velocity. Example 1 A car travels 50 km due East and a further 50 km due South in a total time of 1.0 h. Determine (a) the total distance travelled, (b) the displacement, (c) the average speed, and (d) the average velocity, of the car. Solution: (a) total distance = 50 + 50 = 100 km (b) displacement = 22 50 + 50 = 71 km tan θ = 50 50 = 1.0  θ = 45o displacement is 71 km at a bearing of 135o or S45oE (c) -1 -1100 km h 27total distance 100 total time m1.0 .8 sspeed ==== (d) 1 -1total displacement 71 71 km h 19.7 m stotal time 1.0velocity −= = = = at a bearing of 135o or S45oE N N N

4 2.3 Acceleration (a) Acceleration is the rate of change of velocity. SI unit : m s-2. It is a vector quantity. * A change in velocity can be due to: 1. a change in magnitude – for example: the speed changes while the direction remains the same. 2. a change in direction – for example: an object moving with constant speed in a circle. 3. a change in both magnitude and direction simultaneously. * Instantaneous acceleration is the rate of change of velocity at that instant of time. a = dt dv * Average acceleration = change in velocity time taken v v u tt −== 2.3.1 Sign of Acceleration In describing acceleration, it is necessary to consider the object’s direction of motion and whether it is speeding up or slowing down. * If an object is speeding, the magnitude of velocity increases with time. Acceleration and velocity vectors must act in the same direction. * If an object is slowing down or decelerating, the magnitude of velocity decreases with time. Acceleration and velocity vectors must act in opposite directions. * Negative acceleration does not necessarily mean an object is slowing down. If both acceleration and velocity are negative, the object is speeding up. 3 Graphical Representation of Motion LO(b), (c) The relationships between the three graphs can be summarised Graph Gradient at a point Area under the graph displacement-time, s-t ds dt = instantaneous velocity - velocity-time, v-t dv dt = instantaneous acceleration change in displacement acceleration-time, a-t - change in velocity

5 3.1 Displacement-time graph dis The displacement-time graph gives the following information: 1. Slope or gradient at A = s t   = v, the instantaneous velocity at A 2. Average velocity between O and A, <v> = 1s t Example 2 Consider a body moving along a straight line. Its displacement-time (s-t) graph is as shown. Describe the motion of the body from O to C. Solution: O to A s increases at a constant rate (constant gradient)  body is moving with constant velocity. A to B s is constant (zero gradient)  body is stationary or velocity is zero. B to C s decreases at a constant rate (constant negative gradient)  velocity is negative, meaning body is reversing/moving back towards O. Velocity is constant but greater in magnitude than that of OA. A B C O s t gradient = ors ds t dt   = instantaneous velocity s1 A O displacement s time t t s t

6 (a) (b) (c) (d) (e) 3.2 Velocity-time graph The velocity-time graph gives the following information: 1. Slope or gradient at any time = v t   = a, the instantaneous acceleration 2. Area u