Copyright © 2015 Chris J Jewell 1 Mechanics M1 (Slide Set 4) Linear Motion (Constant Acceleration) Mechanics M1

Copyright © 2015 Chris J Jewell 2 Distance versus Displacement 0 Distance Displacement Distance is a scalar Displacement is a vector θ Mechanics M1

Copyright © 2015 Chris J Jewell 3 Distance versus Displacement 0 Displacement Displacement is a vector time 10 km 1hr 2hr 3hr Start Stop + Vel. -Vel. Resting From the start velocity is positive for 1 hour at v = 10 km hr -1, then zero and finally back to the start location again at v = -10 km hr -1 We get the same magnitude but no direction information, so we have speed not velocity Mechanics M1 Distance Distance is a scalar 0 1hr 2hr 3hr 10 km 20 km Resting

Copyright © 2015 Chris J Jewell 4 Velocity-Time Graphs Area A1 + A2 + A3 = Distance Travelled Area under the acceleration graphs represent change in Velocity A1 = 4 x 15 = 60 ms -1 A3 = - 6 x 10 = - 60 ms -1 Acceleration -Time Graphs v ms t sec a ms -2 t sec A1 A2 A3 A1 A3 Constant velocity = Zero Acceleration Mechanics M1

Copyright © 2015 Chris J Jewell 5 Motion Graph Examples Example 1: A bus starts from stop A and accelerates for 20 minutes at 2.5 mmin -2 until it reaches a constant velocity of 50 mmin -1 After 30 minutes it decelerates at 2 mmin -2 to a stop at B. Draw a velocity – time graph of the motion, calculate the time of arrival and the distance travelled by the train. In triangle A3 Total distance travelled by the bus is Area A1 + A2 + A3 (b) (c) Mechanics M1 Velocity-Time Graphs v ms t min A1 A2 A3 tBtB B A (a)

Copyright © 2015 Chris J Jewell 6 Motion Graph Example 2 - A racing car passes the pits at a steady 60 ms -1 and a second car, initially at rest, joins the track with a constant acceleration of 25 ms -2 for 3 seconds until it reaches a constant velocity of 75 ms -1. Calculate the time the second car catches the first car and the distance travelled at that time. The cars are in the same location when the graph area for each car is the same v ms -1 t sec 0 T 3 That occurs when Total distance travelled is then 60 ms -1 x 7.5 sec = 450 m Mechanics M1

Copyright © 2015 Chris J Jewell 7 Equations of Motion Since this is a velocity – time graph we can also find the distance travelled in the time interval from the graph Distance travelled s = the area under the line = the area of the trapezium ABCD This can be split into the square ABED and triangle DEC to give: 0 u v Velocity Time Final Velocity Initial Velocity Time Interval (t) tAtA tBtB A B D C E Mechanics M1

Copyright © 2015 Chris J Jewell 8 SUMMARY Where “a” is acceleration (ms -2 ), “s” is the displacement (m), “t” is time (s), “u” is initial velocity (ms -1) and “v” is the final velocity (ms -1 ) – these are also referred to as “SUVAT” equations. Newton’s Laws of Motion Mechanics M1

Copyright © 2015 Chris J Jewell 9 Example 1: A particle starts at a point when t = 0, passes point A after 5 sec and point B after 6.8 sec, if the car has a constant acceleration 1.5 ms -2. If the initial speed of the car is 10 ms -1, find (a) the distance AB and (b) the velocity after 10 sec? In this case we have u = 10 ms -1, v = ? ms -1 a = ? ms -2, Time from the start to point A is 5 seconds A B (a) (b) Mechanics M1 Straight Line Motion Examples t = 0 t = 5 sec t = 6.8 sec

Copyright © 2015 Chris J Jewell 10 Example 2: A particle passes point A with a speed of 15 ms -1 and point B with a speed of 30 ms -1. If the particle has a constant acceleration, find the speed of the particle at the midway point between A and B. A B M Moving towards the right, using Moving towards the left, using Adding equations (i) and (ii) eliminates 2as Mechanics M1 Positive 15 ms ms -1

Copyright © 2015 Chris J Jewell 11 Examination Type Questions A train moves with constant acceleration along a straight horizontal track. It passes point A with a speed of 6 ms -1 and 10 seconds later it passes point B. If AB =100 m find (a) the acceleration of the train (b) the speed of the train when it passes point C which is 120 m further down the track,. A 6 ms -1 B ? ms m C 120 m ? ms -1 We know u = 6 ms -1 at A, v = ?, t = 10 sec and AB = s = 100 m. (a) (b) Mechanics M1

Copyright © 2015 Chris J Jewell 12 M1 Mechanics Vertical Motion Mechanics M1

Copyright © 2015 Chris J Jewell 13 Mechanics M1 Vertical motion under gravity. The equations of motion previously described can be applied directly to vertical motion with the acceleration term being always the value due to gravity. It is very important to realise that the equations of motion can be used to evaluate positive and negative values of velocity and displacement, as long as the direction of motion is specified relative to a reference level: Positive up, Negative down. Applications of the velocity formula v = u + at can be illustrated by considering a particle being projected vertically upward with a velocity of 30 m s -1, say, and plotting the results as a velocity – time graph (assuming g = -10 m s -2 for convenience) The value of v = u + gt is shown opposite (red spheres) for t = 0 to t = 6 seconds, defining positive as upward. The initial velocity of u = 30 m s -1 is gradually reduced to v = 0 after 3 s, then increases to – 30 m s -1 at time t = 6 s. Continued on next slide v (m s -1 ) t (s) 30 3 Velocity -Time Graph 45 m m Positive

Copyright © 2015 Chris J Jewell 14 Mechanics M1 Continued on next slide ( 1) As the particle rises its velocity decreases to zero – which occurs at its maximum height above its start point: (2) When the particle returns to its start point its speed is the same as its initial speed, but in the opposite direction – the motion is symmetrical. (3) The total distance of the flight is twice distance to the highest point. Vertical motion under gravity. Notes:

Copyright © 2015 Chris J Jewell 15 Mechanics M1 Continued on next slide These considerations are analysed in more detail below related to the particle projected at 30 m s -1 : Vertical motion under gravity. (5) The slope of the graph (on the previous slide) is – 10 m s- 2 (6) The area under the velocity-time graph gives the total distance travelled. (4) Total travel time can be found using v = u + at, with final velocity = - u

Copyright © 2015 Chris J Jewell 16 Mechanics M1 Displacement The formula for displacement can be written as: This equation completely describes the upward and downward motion of a body under the effect of gravity near the surface of the Earth (excluding any other effects such as air resistance). The nature of this quadratic equation is illustrated with a displacement - time graph below, g has been taken as -10 m s -1 : Note that the axes are displacement and time – But the particle motion is vertically up and down. The motion is symmetrical with the particle exactly repeating each position. Continued on next slide t (s) s (m) Displacement-Time Graph Positive

Copyright © 2015 Chris J Jewell 17 Mechanics M1 Calculation Examples (a) To find out the time when the particle returns to the ground the equation is used as follows: (b) To find the time when the particle, projected from the ground is at 25 m from the start position (+ 25 m above the ground) the equation is used as follows: t (s) s (m) Displacement-Time Graph

Copyright © 2015 Chris J Jewell 18 Mechanics M1 Calculation Examples (c) To find out the time that the particle hits the ground if the particle’s start position is 25m above the ground the equation is used as follows: t (s) s (m) Displacement-Time Graph 25 m Particle thrown from here (Reference Zero) D – T Curve for Particle thrown from 25 m above the ground Displacement to ground is – 25 m Positive

Copyright © 2015 Chris J Jewell 19 Mechanics M1 Further Example Calculations A ball projected upwards with a velocity of 13 m s -1 from a height x above the ground – Neglecting air resistance work out x if the ball takes 3 s to reach the ground. Neglecting air resistance work out the velocity of the ball when it hits the ground. Reference Zero - x 13 m s m s -1 Positive

Copyright © 2015 Chris J Jewell 20 Motion under Gravity – Acceleration is always downwards at 9.8 ms -2 Example 1 : A stone is dropped into a well and a splash sound is heard 3.5 seconds later, how deep is the well? We have t = 3.5 sec, u = 0 ms -1, v = 0 ms -1 a = 9.8 ms -2 and we need to find s,, so we can use: s Mechanics M1 Positive

Copyright © 2015 Chris J Jewell 21 Example 2 : A tennis ball thrown vertically from a window 32 m above the ground with a velocity of 16 ms -1, find (a) the maximum height above the ground attained by the ball (b) the time that the ball touches the ground? (a) Maximum height: From v 2 = u 2 + 2as, since v = 0 at the highest point: -32 m +ve So the maximum height is 32 m m = 41.5 m Not to scale: Zero Level Maximum Height m Mechanics M1 (b) Time to reach the ground

Copyright © 2015 Chris J Jewell 22 Examination Type Questions A ball is projected upwards from the ground with a velocity of 24 ms -1. Find (a) its maximum height (b) the times that the ball is more than 24 metre above the ground. (a) Using v 2 = u 2 + 2as the highest point is reached when v = 0 i.e. when 0 = 576 – 19.6s or s = 576 / 19.6 = 29.4 m Maximum Height 24 m Ground (b) (c) Time above 25 m = 3.5 – 1.4 = 2.1 sec Mechanics M1