EJC Physics H202 Kinematics 2023 1. Notes (FULL)

Content • Rectilinear motion • Non-linear motion Learning Outcomes Candidates should be able to: (a) show an understanding of and use the terms 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 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 accelera ted motion in a straight line (e) solve problems using equations which represent uniformly accelerated motion in a straight line, including the motion of bodies falling in a uniform gravitational field without air resistance (f) describe qualitatively the motion of bodies falling in a uniform gravitational field with air resistance (g) describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction. Kinematics describe the motion of points, bodies and systems of bodies without considering the forces that cause them to move. Before we can fully appreciate the motion, we need a common vocabulary to help establish an understanding of the science behind describing motion. Kinematics is often referred to as the “geometry of motion” as the values of position, displacement, velocity and acceleration are worked on mathematically without considering the type of force causing the motion. We shall talk about how forces act on bodies in the topic of Dynamics. “Perseverance” Mars rover landing. The intense entry, descent, and landing (EDL) phase begins when the spacecraft reaches the top of the Martian atmosphere, traveling at nearly 20 000 kmph. Many refer to the time it takes to land on Mars as the “seven minutes of terror.”

We typically denote position vectors using s with respect to an origin. Then the displacement (or change in displacement) can be final initials s s = − . Example 1 A car travels on a straight horizontal road. It starts moving at time t = t1 and comes to a stop momentarily at t = t2. It then reverses and comes to a stop again at t = t3. Taking rightwards direction as positive, state the (a) total distance travelled, and (b) displacement of the car with respect to origin x = 0. –30 –20 –10 0 10 20 30 40 x/m distance / m t = t1 t = t2 t = t3 time distance travelled displacement from origin t1 0 m 0 m t2 30 m +30 m or 30 m to the right t3 90 m –30 m or 30 m to the left y x Note: It is quite conventional to take rightwards as positive for horizontal motion and upwards as positive for vertical motion. It is equally conventional to take the direction normal and away from an inclined surface as positive displacement from the surface. It is always clearer to include a sketch or a phrase of how you define your convention. Distance is the total length of the actual path travelled between the start and the finish points. Displacement is the straight line distance from the start to the finish point in that direction. displacement O distance

Example 2 From point A, a car travelled 300 m towards the east to point B. It then travelled 400 m north to point C. Calculate the total distance it travelled and its displacement from its starting position. Note: Reminder to give both magnitude and direction for vector quantities such as displacement. Speed is a scalar quantity and so has magnitude only. Velocity is a vector quantity and so has magnitude and direction. Both have units of m s-1. displacement of car is [magnitude] 500 m, [direction] 53.1 north of east (or bearing of 36.9°) C N 400 m A 300 m B ϴ t1 displacement s t ∆t ∆s sf si t2 average speed = average velocity The gradient of the tangent to the graph at t1 gives the instantaneous velocity. When we refer simply to velocity, we generally mean the (instantaneous) velocity at that point in time. t1 s t Speed is the rate of change of distance. Velocity is the rate of change of displacement. = dsv dt s : displacement (m) t : time (s) v : velocity (m s-1)

Example 3 Cars A and B are travelling in the directions as shown. The speeds and velocities of the cars are given. Note: The velocity of A relative to B is ( ) 1 AB A B A B 7 m s to the righ0 tv v v v v −= − = + − = . Acceleration is a vector quantity and has direction. Deceleration refers to decreasing speed, regardless of direction. Average acceleration can be found by –30 –20 –10 0 10 20 30 40 x/m distance / m speed of A = 40 m s-1 speed of B = 30 m s-1 car speed velocity A 40 m s-1 + 40 m s-1/ 40 m s-1 to the right / 40 m s-1 in positive direction B 30 m s-1 -30 m s-1/ 30 m s-1 to the left / 30 m s-1 in negative direction average acceleration final initialchange in velocity time taken t va v−= = Acceleration is the rate of change of velocity. = dva dt v : velocity (m s-1) t : time (s) a : acceleration (m s-2)

speed up while moving in positive direction slow down while moving in positive direction final initial final initial so is positive v v t v va  − = final initial final initial so is negative v v t v va  − = (car is decelerating but a is negative) speed up while moving in negative direction slow down while moving in negative direction ( ) ( ) final initial final initial so 40 30 is negative v va t v v t  −  − − − = =  ( ) ( ) final initial final initial so 30 40 is positive v va t v v t  −  − − − = =  (car is decelerating but a is positive) The car speeds up whenever acceleration is in the same direction as the velocity. It slows down when its acceleration is opposite to its velocity. vi = 30 ms–1 vf = 40 ms–1 a vi = 40 ms–1 vf = 30 ms–1 a vf = - 40 ms–1 vi = - 30 ms–1 a vf = - 30 ms–1 vi = - 40 ms–1 a

Example 4 An athlete sprints for 60.0 s at a constant speed round a 400 m track in an anti-clockwise direction. (a) Determine his (i) average speed for the whole journey, (ii) average velocity for the whole journey, (iii) average velocity from A to C, (iv) instantaneous velocity at D, (v) acceleration at the midpoint of B and C, (vi) average acceleration for the whole journey. (b) Is the athlete accelerating when he is halfway between C and D? 100 m 100 m A B 63.7 m C D Solution: 3.97 m s-1 at 57.5 anticlockwise of AB along DA (a)(v) no change in velocity so 0 acceleration (a)(vi) no change in velocity so 0 acceleration (b) [magnitude] while speed is constant [direction] direction of motion is constantly changing between C and D so velocity, being a vector, is changing so he is accelerating 63.7 m 100 m A θ B C

Motions graphs track the change in s, v, and a throughout time. Graphs are therefore very rich in the information they can contain and present: type of graph gradient represents area represents displacement – time s-t velocity d d=v s t (nil) velocity – time v-t acceleration d d= va t change in displacement so d dd tsv st v==  acceleration-time a-t (nil) change in velocity so d dd tva vt a==  Example 5 In the displacement-time graph as shown, determine the (a) average speed from t = 0.0 s to t = 4.0 s, (b) average velocity from t = 0.0 s to t = 4.0 s, (c) instantaneous velocity at t = 3.5 s and (d) instantaneous speed at t = 3.5 s. Solution: We can