19.Kinematics

19. KINEMATICS

Kinematics is the branch of mathematics that deals So, we define velocity as the rate of change of with the motion of a particle in relation to time. In displacement. this topic, we are concerned with quantities such as displacement (and distance), velocity (and speed) and Displacement Velocity= acceleration. In studying the motion of objects, we Time will examine travel graphs, laws of motion and the application of calculus to problems in kinematics. When the velocity (or speed) of a moving object is increasing the object is accelerating. If the velocity Speed, distance and time or speed decreases it is said to be decelerating. When a particle moves at a constant speed, its rate of Acceleration is, therefore, the rate of change of change of distance remains the same. Constant speed velocity. Constant acceleration is calculated from the is calculated from the basic formula formula

Distance Velocity Speed = Acceleration = Time Time

If speed changes in an interval, the average speed When speed is constant, there is no acceleration, that during the period of travel can be calculated from the is, acceleration is equal to zero. formula: Since acceleration is velocity per unit time, units for acceleration will take the form such as ms-2, cms-2, Total distance covered or travelled kmh-2. For example: Average Speed = −1 Total time taken Velocity ms 1 1 2 Acceleration = = = ms− − = ms− Time s Units In calculations involving speed, distance and time, Example 2 units must be consistent. So, if the units in a given A car starts from rest and within 10 seconds is situation are not consistent, then we ought to change moving at a velocity of 20ms-1. What is its the units to successfully solve the problem. acceleration, given that the acceleration is constant? Example 1 If a car travels at a speed of 10ms-1 for 3 minutes, Solution calculate the distance it covers. At rest, velocity = 0 ms-1 Velocity after 10 s = 20 ms-1 Solution To solve this problem, we must change the 3 20 0 ms-1 Change in velocity ()− -2 minutes to 180 seconds, because speed is given in Acceleration = = = 2ms Time 10 s metres per secondwww.faspassmaths.com. Using, Distance = speed × time. Example 3 Distance travelled = 10 × 180 =1800 m=1.8 km A particle moving in a straight line at 18 ms-1

decelerates at a constant rate, reaching a speed of 6 -1 Velocity and acceleration ms after a period of 24 seconds. Velocity is the speed of a particle measured in a Calculate the deceleration. specified direction of motion. Therefore, velocity is a vector quantity, whereas speed is a scalar quantity.

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 1 of 9 Solution The different segments of the journey are all Initial speed = 18 ms-1 Final speed = 6 ms-1 displayed using straight lines. Consider his starting point as home. From the graph we can deduce: Change in velocity Acceleration = i. The cyclist left home at 8 a.m. and returned Time taken home at 5 p.m. (the ending point on the graph). Final velocity-Initial velocity Note that at 8 a.m. and 5 p.m., he was zero = Time taken distance away from home. 6 18 ms-1 ii. At 9 a.m. he was 10 km away from home and at ()− 1 -2 = = − ms 10 a.m. he was 20 km away from home. Also, 24 s 2 between 2 p.m. and 3 p.m. he was 60 km away (The negative sign indicates that the particle is from home. This was the maximum distance decelerating) travelled. iii. At 3 p.m., he began his return journey and 1 \Deceleration is ms-2 between 3 p.m. and 5 p.m., he was travelling 2 towards home. iv. Between 10 and 11 a.m. and between 2 and 3 Travel graphs p.m. the cyclist was at rest. Horizontal lines on Travel graphs display the motion of a particle over a the graph for these periods indicate that there is period of time. Some of the simplest kinds are graphs no motion. of distance (or displacement) versus time and v. The cyclist travelled at a constant speed during velocity (or speed) versus time. On travel graphs, the periods 8 a.m. to 10 a.m., 11 a.m. to 2 p.m., time always appears on the horizontal axis because it and 3 p.m. to 5 p.m. is the independent variable. vi. The speed during each of the 3 segments is calculated from the gradient of the lines as (a) Displacement/Distance-time graphs shown below. In these graphs, the vertical axis displays the distance from a certain point and the horizontal axis displays Speed from 8 a.m.to 10 a.m the time. When a body is moving at constant speed, a 20 km straight line distance-time graph is obtained. The = 2 hr journey of a cyclist is shown below. =10kmh-1

Speed from 11 a.m.to 2 p.m 40 km = 3 hr =13.3 kmh-1 Speed from 3 p.m.to 5 p.m 60 km = 2 hr −1 www.faspassmaths.com = 30 kmh

If the speed or velocity of an object is not constant, the distance/displacement-time graph will be represented by a curve. From such graphs, we calculate the speed/velocity at an instant by finding the gradient of a curve at a point. This is done by drawing a tangent at that point and finding the

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 2 of 9 gradient. From a graph of distance/displacement Rise 15 m = against time we can obtain: Run 5 s i. The distance covered at any time by a read = 3ms-1 off.

ii. The time taken to cover any distance by a (b) Velocity/Speed -time graphs read off, and A speed/velocity-time graph displays the velocity or iii. the speed/velocity at a given point by speed of the particle on the vertical axis and time on calculating the gradient at that point on the the horizontal axis. For a body moving with uniform graph. acceleration, a straight-line velocity-time graph is

obtained. From such a graph, we can deduce that: When speed/velocity is not constant, we obtain a curved distance/time graph as shown below. 1. A horizontal line indicates that the object is

at rest, that is, the velocity is zero. 2. The gradient gives the constant acceleration or deceleration. A positive slope indicates acceleration while a negative slope indicates deceleration 3. The area under the graph is the distance covered.

The graph below shows a straight-line velocity-time graph displaying the journey of two cyclists. The first cyclist (blue line) travelled at constant acceleration throughout the journey. The second cyclist travelled at constant acceleration then stopped accelerating and

travelled at a constant velocity. Afterwards, the In the graph shown, cyclist travelled at constant deceleration and finally (i) the distance is s , at the time, t . 1 1 came to a stop.

(ii) the time taken to cover a distance of s2 is t2 (these are obtained by a simple read off)

(iii) the gradient of the curve at P is the speed at t3 . Example 4 The graph shows the distance-time curve for a particle. Find the speed at time, t.

Solution www.faspassmaths.com

We can make the following deductions: i. The acceleration of the first cyclist was The speed at time t is the gradient of the curve at t 10�� = 5�� 2�

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 3 of 9 ii. The acceleration of the second cyclist was Area of A4 =41612()-= 16 8�� = 2�� \Total area =+++=16 48 12 16 92 4� The distance covered is 92 m. iii. Between 4-7 seconds the second cyclist travelled For a body moving with non-uniform or variable at a constant velocity. acceleration, a curved velocity-time graph is obtained iv. The deceleration of the second cyclist was as shown below. 8ms-1 = 2.67 ms-2 (to 2 d.p) 3s v. The total distance travelled by the second cyclist is found by the area of the trapezium 1 =´8(10+ 3) metres= 52 metres 2

Example 5 A velocity/time graph is shown below.

From this graph, we can deduce the following:

1. The velocity at any time or the time taken to attain any velocity can be obtained by a read-off.

For example, it took a time of t2 to attain a Calculate: velocity of v2 . a. the acceleration from to t = 10 t = 12 2. The area under the curve gives the total distance b. the distance covered in the first 16 seconds covered. For example, the area of the shaded region gives the total distance covered between Solution t and t . a. The acceleration during the period t = 10 3 4 3. The gradient of the tangent at a point gives the to t = 12 is the gradient of the line joining acceleration at that point. For example, the 84- 4 (10, 8) to (12, 4) ===-2 gradient of the tangent at P gives the acceleration 10-- 12 2 -2 t Acceleration = –2 ms (the negative sign at 5 . indicates a deceleration) Example 6 b. The areawww.faspassmaths.com under the graph will give the The velocity, v ms-1, of a ball at time t is given by distance covered. The region is divided vtt=-0.52 + 4 + 6 . into 4 regions, as shown in the diagram i. On graph paper draw the velocity-time graph above. for the first 10 seconds. 48´ ii. Determine the velocity of the ball after 2 Area of A ==16 1 2 seconds. iii. Estimate the acceleration of the ball after 6 Area of A2 =8104´() - = 48 seconds. 1 Area of A =+()()84´- 121012= 3 2

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 4 of 9 Solution Notation The following symbols are used to represent the (i). The velocity-time graph for the first 10 variables we are interested in. seconds is shown in the diagram shown below. u = initial velocity v = final velocity a = acceleration t = time s = displacement from a fixed point at time, t = 0 .

First Law By definition, Change in velocity Constant Acceleration = Time vu- \avua= or -=t (1) t

Second law (ii). The velocity after 2 seconds can be By definition, for constant acceleration obtained by a read-off from the graph. As Displacement = average velocity ´ time shown on the graph, the velocity at t = 2 is æöuv+ st=ç÷´ 12 ms-1. èø2 (iii). To obtain acceleration at t = 6, we draw the Replace v by u+ at tangent at this point and find the gradient æöæöu++ u at2 u + at of the tangent. We calculate, stt=ç÷ç÷´=´= (� + ��) × � èøèø22 Rise 5 ==-2ms-2 . 1 Run- 2.5 sutat=+ 2 (2) 2 Hence, the acceleration at t = 6, is estimated as Third law -2 ms-2. If we square the first equation:

()v=+ u at 2

22 222 æö1 2 v=+ u22 uat + a t =+ u aç÷ ut + at èø2 1 Recall: sutat=+ 2 2 \v22= u +2 as (3)

The equations (1), (2) and (3) are called the three www.faspassmaths.comequations of linear motion and may be used in calculations where there is linear motion under a constant acceleration.

The equations of linear motion are: The motion of a particle in a straight line 1. � = � + �� under constant acceleration 2. � = �� + �� For linear motion under constant acceleration, we use 3. � = � + 2�� equations of motion to calculate unknown quantities.

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 5 of 9 Example 7 (iii) From B to the rest position. A particle starts from a point O and accelerates u = initial velocity = 24 ms-1 v = final velocity = 0 uniformly for 8 seconds at a constant rate of ms-1 and a = − 3 ms-2 2 ms-2, until it reaches a point P. Calculate: v=+ u at i. the velocity at P. 0=+ 24()- 3 t

ii. the distance covered. 3t = 24

t = 8s Solution From B, the particle takes 8 s to come to rest. i Let u = initial velocity = 0 ms-1 (at O), t = 8 s v = final velocity or velocity at P. Acceleration, a = 2 ms-2 Using calculus to solve Kinematics Problems vuat=+ We can use differentiation and integration to solve problems in kinematics. We use the ideas of rate of v =+02816ms = -1 () change to obtain the relationships. The velocity at P is 16 ms-1 Notation ii Let s = distance covered, then The following symbols are used to represent the 1 s=+ ut at 2 variables in kinematics. 2 1 2 =+08() ()() 2 8 2 v = velocity t = time a = acceleration = 64 Distance covered = 64 m. x = displacement from a fixed point at time, t = 0 .

Example 8 A particle passes a fixed point, A, and decelerates Recall that velocity is the rate of change of distance -2 uniformly at 3 ms for 12 s until it reaches a point (� = ), and acceleration is the rate of change of -1 B. The velocity at B is 24 ms . velocity (� = ). From these two relationships, we i. Calculate the velocity at A. ii. Calculate the distance from A to B. can use calculus to derive other relationships as iii. From B, how long will the particle take to shown below. come to rest? dx dv v = a = Solution dt dt (i) Let velocity at A be u ms-1 and acceleration, the dx= vdt dv= adt

deceleration, � = −3��, time, t = 12 s and the dx= vdt dv= adt -1 òò òò final velocity (or velocity at B), v = 24 ms v=+ u at xvdt= ò vadt= ò \24=+u ()()- 3 12 dx v = u =+24 36www.faspassmaths.comdt -1 d dæö dx = 60 ms ()v = ç÷ dt dt dt èø dv d2 x (ii) Let the distance from A to B be s = 2 s= �� + �� dt dt 2 dx s= 60(12) + (−3)(12) a = dt 2 � = 720 ∗ (−216) � = 504 �

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 6 of 9 Summary Example 10 It is important to note that in using calculus to solve A particle starts to move at O and travels in a problems in kinematics we must be given an straight line with velocity v cms-1, where expression for x or v or a, in terms of t. We then vtt=2 -2 and t is the time in seconds after leaving choose to integrate or differentiate, with respect to t, O. The particle comes to rest at A. Calculate to solve for the required unknown quantity. a. the value of t at A. b. the acceleration, a cms-2 at A. �� = � �� = � c. the distance, s cm, from O to A. �� d. the total distance covered from t = 0 to t = 4 . � �� = = � � �� = � �� Solution

a. vtt=2 -2 , At A, particle is at rest, sov = 0 2 20tt-= Example 9 tt20-= The velocity, v ms-1, of a particle moving from O, () in a straight line, is given by vt=6532 - t+, t = 0 or t = 2 where t is the time in seconds after leaving O. t = 0 at O and Calculate t = 2 at A. a. the initial acceleration of the particle b. the distance from O when t = 4 . b. The acceleration at A vtt=2 -2 Solution dv a = , \at=22- a. The initial acceleration is the acceleration when dt t = 0. At A, a =222-() =- 2cms-2. vt=6532 - t+ Let the acceleration at time t be a. dv c. The distance from O to A att==62() - 5= 12- 5 dt svdt= Initial acceleration is at t = 0, ò a = 12(0) −7 =-5 ms-2. t3 sttdttC=()2 -22=-+ ò 3 b. Let the distance from O at time t be s. s = 0 at t = 0 65tt32 svdtt==6532 - t+= dt-++ 3 tC, 3 òò() 2 ()0 32 \sC=()0 - + , C = 0 where C is a constant 3 3 3 325 2 t 2 ()2 1 \st=23- t+ tC + \st= - , s =()21-= 2 3 33 s = 0 at t = 0 (data) d. The total distance covered from t = 0 to t = 4 . 325 \sC=20- 0+ 30 + From part (a), we note that the particle was at rest () ()() www.faspassmaths.com2 at � = 2 sec. \C =0 Total distance covered from 2 to 4 seconds is 325 st=23- t+ t � 2 (2� − �) �� = � − 3 325 s =24() -()() 4+= 34 100 2 = 16 − − 4 − = −5 − 1 = −6 cm Distance from O when t = 4 is 100 m.

Total distance covered = 1 + 6 = 8 ��

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 7 of 9 Example 11 � = 3� − 4� + 6 A particle passes through a fixed point O and at a �� = = 9� − 4 time t seconds after leaving O, its velocity is v �� cms-1 is given by vptqt=++3 6 where p and q When � = 0, 9� − 4 = 0 are constants. It is also given that the distance, s 2 2 t =± t > 0 \t = only cm, from O at time t = 2 , is 16 cm and the 3 3 acceleration, a cms-2, is 32 cms-2. The velocity when � = 0, Calculate 3 a. the value of p and of q. æö22 æö v =346ç÷- ç÷+ b. the velocity at the time when a = 0 . èø33 èø

88 v =-+6 Solution 93 2 a. vptqt=++3 6 � = 4 �� 9 s==òò v dt() pt3 ++ qt6 dt Example 12 pt42 qt st=+++6 C, C is a constant A particle, P, moves in a straight line so that its 42 displacement, s cm, from O, at time t seconds after s = 0 when t = 0 leaving O, is given by sttt=+28 4-- 5 23cm. 42 pq()00() Another particle, Q, moves in the same straight \06= + +()0 +C 42 line as P and starts from O at the same time as P. \C =0 The initial velocity of Q is 2 cms-1 and its -2 pq acceleration is a cms given by at=26- . \st=42 + tt +6 42 Find t when P and Q collide and determine s = 16 when t = 2 whether P and Q are travelling in the same or opposite direction at the instant of collision. pq2242 () () \16 = + +6() 2 42 Solution 424pq+= For particle Q 22pq+=…(1) at=26- \vtdt=()26- ò 2 a = 32 when t = 2 = - 32 + Cttv Since the initial velocity of Q is 2 cms-1,when dv 2 aptq==´3 + � = 2, � = 0 dt 2 and 32=+ 3pq 2 2 \ = ()()- 03022 + C () 12pq+= 32 …(2) \C = 2 2 vtt=23- + 2 Equation (2) – Equation (1) Let s be the distance travelled by Q after t 12pq+= 32 www.faspassmaths.com seconds, 22pq+= s = vdt = 2t - 3t 2 + 2 dt ò ò() 10p = 30 andp = 3 23tt23 stK=-++2 From (1) � = 2 − 2(3) = −4 23 \p =3 andq =-4 32 = - 2 ++ Kttts When � = 0, � = 0 b. The velocity when a = 0 \00=()()()23- 0+ 20 +K, Substituting the values of � and � in �: Hence, � = 0

At the point of collision, both particles would have travelled the same distance from O.

\28 + 54 -- = - 3232 + 2tttttt - 2 26 tt 28 =++ 0 2 26 tt -- 28 = 0 3 2 tt -- 14 = 0

(- )(tt ) =+ 0273 1 1 tt==2 or - 2 Since t > 0, t = 2 only 3 3

For P sttt=+28 4-- 5 23 ds \vtt= =4103-- 2 dt 2 æö77 æö When the velocity of P v =410--ç÷ 3 ç÷ èø33 èø 11 =42316-- = negative 33

For Q 2 1 æö77 æö When t = 2 v =23ç÷- ç÷+ 2 3 èø33 èø 21 =4162- + 33 21 =4162- += negative 33

\At the point of collision, both P and Q are travelling in the same direction.