Motion in One Dimension Lecture Notes PDF

Motion in one dimension Dr/ Mai Hosny By Dr/Mai Hosny  Position, displacement and distance  Position The line is coordinated and referenced from a point O, the origin. For a horizontal line, the convention is that positions to the right of O are positive...

Motion in one dimension Dr/ Mai Hosny By Dr/Mai Hosny  Position, displacement and distance  Position The line is coordinated and referenced from a point O, the origin. For a horizontal line, the convention is that positions to the right of O are positive, and positions to the left are negative. Dr/ Mai Hosny  The position of a particle is often thought of as a function of time, and we write x(t) for the position of the particle at time t.   Displacement The displacement of a particle moving in a straight line is the change in its position. If the particle moves from the position x(t1) to the position x(t2), then its displacement is x(t2) − x(t1) over the time interval [t1,t2]. In particular, the position of a particle is its displacement from the origin. Dr/ Mai Hosny If a particle moves from O to B, its displacement is 3 m. If a particle moves from O to A, its displacement is −4 m. If a particle moves from A to B, its displacement is 7 m. If a particle moves from B to A, its displacement is −7 m.  Distance The distance is the ‘actual distance’ travelled. Distances are always positive or zero. For example, given the following diagram, if a particle moves from A to B and then to O, the displacement of the particle is 4 m, but the distance travelled is 10 m. Dr/ Mai Hosny  Constant velocity The rate of change of the position of a particle with respect to time is called the velocity of the particle. Velocity is a vector quantity, with magnitude and direction. The speed of a particle is the magnitude of its velocity. Dr/ Mai Hosny  For example, consider a particle which starts at the origin O, moves to a point B at a constant velocity, and then moves to a point A at a different constant velocity, as shown in the following diagram. Dr/ Mai Hosny NOTE Dr/ Mai Hosny  Velocity–time graphs  The following diagram shows the velocity–time graph for a particle moving at 5 m/s for 4 seconds. The constant velocity is plotted against time. Naturally, it gives a line segment parallel to the t-axis. The gradient of this line is zero. Dr/ Mai Hosny  area represents the displacement.  Position–time graphs Positive velocity Assume that a particle moves with constant velocity from point A to point B, as shown in the following diagram. At time t = 1, the position of the particle is 3 m to the right of O, that is, x(1) = 3. At time t = 4, its position is 9 m to the right of O, that is, x(4) = 9. Dr/ Mai Hosny The displacement (change of position) of the particle is 6 meters over the time interval [1, 4]. The duration of the motion is 3 seconds. Therefore the constant velocity is 2 m/s. The position–time graph for this motion is as follows.  Dr/ Mai Hosny  the gradient of the line is equal to the velocity.  Negative velocity Assume that a particle moves with constant velocity from A to B, as in the following diagram. At time t = 1, the position of the particle is 3 m to the right of O, that is, x(1) = 3. At time t = 4, its position is 3 m to the left of O, that is, x(4) = −3. Dr/ Mai Hosny  The displacement is −6 meters over 3 seconds. Therefore the constant velocity is −2 m/s. The gradient of the position–time graph, shown below, is Dr/ Mai Hosny Dr/ Mai Hosny  Constant acceleration  If the velocity of the particle changes at a constant rate, then this rate is called the constant acceleration. Example Let t be the time in seconds from the beginning Dr/ Mai Hosny of the motion of a particle. If the particle has a velocity of 4 m/s initially (at t = 0) and has a constant acceleration of 2 m/s2 , find the velocity of the particle: 1. when t = 1 2. when t = 2 3. after t seconds. Draw the velocity–time graph for the motion. Dr/ Mai Hosny  The constant-acceleration formulas for motion in a straight line  The five equations of motion Dr/ Mai Hosny or Example A stone is launched vertically upwards from ground level with the initial velocity 30 m/s. Assume that the acceleration is −10 m/s2. 1 Find the time taken for the stone to return to the ground again. 2 Find the maximum height reached by the stone. 3 Find the velocity with which it hits the ground. 4 Sketch the position–time graph and the velocity–time graph for the motion. 5 Find the distance covered by the stone from the launch to when it returns to earth. Dr/ Mai Hosny  Therefore t = 0 or t = 6. Thus the time taken to return to the ground is 6 seconds. Dr/ Mai Hosny Dr/ Mai Hosny Dr/ Mai Hosny