Graphs of Motion Investigation Aim To investigate the relationships between displacement, velocity, acceleration and time for CONSTANT VELOCITY and ACCELERATED motion. Tasks 1. By making use of data logging equipment, produce graphs of displacement vs time, velocity vs time acceleration vs time for motion at CONSTANT VELOCITY and constant ACCELERATION. Describe the relationship observed in each graph in words. Calculate the gradients of the displacement and velocity graphs. Calculate the areas under the acceleration and velocity graphs. Indicate in each case the significance of the gradient and area calculated. Presented in a word document – placed in your hand in folder.

Description in words of what the graph shows Description in words of what the graph shows. You can also use arrows on the graph to make it clear!

Graph Skills y y x y2 - y1 x2 - x1 The ……..…… (……..) of a straight line graph is …………. and can be found using the coordinates between any two points. The gradient of a curve ………………………. continually. At any point the gradient is found by taking the gradient of a ……………….. to the curve at that point. Grad = = y2 2 (x2;y2) y y1 x 1(x1,y1) x x1 x2 y x y2 - y1 x2 - x1 2 (x2;y2) Gradat x = = y 1(x1,y1) x x

Graph Skills y The gradient (slope) of a straight line graph is constant and can be found by taking the coordinates between any two points. The gradient of a curve changes continually. At any point the gradient is found by taking the gradient of a tangent to the curve at that point. y x y2 - y1 x2 - x1 Grad = = y2 2 (x2;y2) y= y2 - y1 y1 1(x1,y1) x = x2 - x1 x x1 x2 y x y2 - y1 x2 - x1 2 (x2;y2) Gradat x = = y= y2 - y1 1(x1,y1) x = x2 - x1 x

Constant Velocity Example Assume you walked for 5s covering 1meter every second. Time (S): 0| 1 | 2 | 3 | 4 | 5 | X (m): 0| 1 | 2 | 3 | 4 | 5 | time (s) Displacement (m) Average Velocity (m/s) Instantaneous Velocity (m/s) Acceleration (m/s2) 1 2 3 4 5

Constant Velocity Example Assume you walked for 5s covering 1meter every second. Time (S): 0| 1 | 2 | 3 | 4 | 5 | S (m): 0| 1 | 2 | 3 | 4 | 5 | time (s) Displacement (m) Average Velocity (m/s) Instantaneous Velocity (m/s) Acceleration (m/s2) 1 2 3 4 5

Constant Velocity Example For each set of data - draw graphs for: displacement vs. time Velocity vs. time Acceleration vs. time x (m) t (s) v (m/s) t (s) a (m/s2) t (s) time (s) Displacement (m) Average Velocity (m/s) Instantaneous Velocity (m/s) Acceleration (m/s2) 1 2 3 4 5

Displacement vs. time - 5 4 3 2 1 y2 - y1 x2 - x1 ... - … … - … y x Grad = = = = … 5 4 3 2 1 X s (m) grad = y2 - y1 x2 x1 5 1 m/s X I I I I I t (s) 0 1 2 3 4 5

Calculating Gradients Choose two points on the line 1 & 2 (as far away from each other as possible) Read off their x & y coordinates (x1;y1) and (x2;y2) Calculate the change in x (x) and y (y): x = x2 - x1 y = y2 - y1 Now work out the gradient using y x point 2: (x2, y2) (5;5) 2  y = y2 - y1 = 5 - 1 1  Point 1: (x1, y1) (1;1) 1 5 x = x2 - x1 = 5 - 1 Gradient = y/x = (5 – 1)/(5 - 1) = 1.0

Displacement vs. time - 5 4 3 2 1 s (m) X X I I I I I t (s) s (m) X X I I I I I t (s) 0 1 2 3 4 5

Displacement vs. time for Constant Velocity - 5 4 3 2 1 X Gradient = y/x = (5-1)/(5-1) = 1 = velocity! (av) X x (m) X Every second the SAME change in displacement takes place. X X I I I I I t (s) 0 1 2 3 4 5

Displacement vs. time S (m) t (s) 5 5 The displacement time graph for constant velocity motion is a ………….line. (Positive or Negative slope) Now calculate the GRADIENT of this graph. GRADIENT = m = /\y/ /\x = …/… = … m/s = ........................! Gradient of displacement vs time graph equals the ………………. of the motion.

Displacement vs. time X (m) t (s) 5 5 The displacement time graph for constant velocity motion is a straight line. (Positive or Negative slope) Now calculate the GRADIENT of this graph. m = /\y/ /\x = 5/5 = 1 m/s = AV. VELOCITY Gradient of displacement vs time graph equals the velocity of the motion.

Velocity vs. time 2 1 X v (m/s) I I I I I t (s) 0 1 2 3 4 5

Velocity vs. time 2 1 v (m/s) X X X X X I I I I I t (s) 0 1 2 3 4 5

Velocity vs. time Area = l x b = 1x5 = 5 (m.s-1 )(s) = m Area = 5m

Velocity vs. time m = /\y/ /\x = …/… = … m/s2 AREA = ………………………….. 1 v (m/s) t (s) 5 Now calculate the GRADIENT and AREA of this graph. m = /\y/ /\x = …/… = … m/s2 Gradient = …………………….. AREA = L x b = …. x …. = ….. m (…..)X…. AREA = …………………………..

Velocity vs. time AREA = l x b = 1 x 5 = 5 m (m/s).s v (m/s) t (s) 5 Now calculate the AREA and GRADIENT of this graph. AREA = l x b = 1 x 5 = 5 m (m/s).s AREA = displacement! m = /\y/ /\x = 0/5 = 0 m/s2 Gradient = acceleration!

DISPLACEMENT - TIME Graph SKETCH GRAPHS - Constant Velocity 1 2 ... . . . . . . . . . . . . . . . . . . . . . . . . . .. . .... S1 S2 Displacement increases by same amount in equal time intervals. time/(s) 0 1 2 3 4 5 6 7 8 9 10 DISPLACEMENT - TIME Graph VELOCITY - TIME Graph s/(m) 2. v/(m/s) . 1. 2. GRADIENT = ……….. . s2 . AREA = ………….. 1. s1 t1 t2 t/(s) t/(s) The ……………… between any two points on the VELOCITY - TIME graph equals the objects displacement between the two points. The ……………………of the DISPLACEMENT vs TIME graph equals the velocity of the object. (at any point)

DISPLACEMENT - TIME Graph SKETCH GRAPHS - Constant Velocity 1 2 ... . . . . . . . . . . . . . . . . . . . . . . . . . .. . .... S1 S2 Displacement increases by same amount in equal time intervals. time/(s) 0 1 2 3 4 5 6 7 8 9 10 DISPLACEMENT - TIME Graph VELOCITY - TIME Graph X /(m) 2. v/(m/s) . 1. 2. GRADIENT = velocity . s2 . AREA = displacement 1. s1 t1 t2 t/(s) t/(s) The area between any two points on the VELOCITY - TIME graph equals the objects displacement between the two points. The gradient of the DISPLACEMENT vs TIME graph equals the velocity of the object. (at any point)

Constant Acceleration Assume you started from rest but traveled further every second. Time (S): 0| 1 | 2 | 3 | 4 | 5 | S (m): 0| 0.5| 1.5 | 2.5 | 3.5 | 4.5 | t(s) X (m) Vav Vi a 1 0.5 2 3 4.5 4 8 5 12.5 Calculate aV -Average velocity Instantaneous velocity Acceleration

If an object accelerates uniformly from rest to 120 km/hr over a certain time interval A – B. (10s) Dx = ( Vi + Vf ) Dt 2 (…..) (…….) ………. m ….. Vi ….. tA tB What is the average velocity?... av V = …/…= …….. = ….. km/hr He will be doing …… km/hr (his average velocity) at ………………….. The average velocity over an interval = the instantaneous velocity at the hallway point.

instantaneous velocity An object accelerates uniformly from rest to 120 km/hr over (12 s) it can be shown it will cover 720 m in that time. Dx = ( Vi + Vf ) Dt 2 (0.0) (120.0) (12.00) 720 m t (s) Vi(m/s) 1 10 2 20 3 30 4 40 5 50 6 60 7 70 8 80 9 90 100 11 110 12 120 120 instantaneous velocity Vi 60 average velocity 0 6 12 What is the average velocity?... av V = x/t = 720/12 = 60 km/hr He will be doing 60 km/hr (his average velocity) at the halfway point!!! The average velocity over an interval = the instantaneous velocity at the hallway point.

Constant Acceleration Assume you started from rest but traveled further every second. Time (S): 0| 1 | 2 | 3 | 4 | 5 | S (m): 0| 0.5| 1.5 | 2.5 | 3.5 | 4.5 | t X (m) aV Vi a 1 0.5 0.5/1 = 0.5 2 2/2=1 3 4.5 4.5/3= 1.5 4 8 8/4=2 5 12.5 12.5/5 = 2.5 Calculate aV -Average velocity Instantaneous velocity Acceleration

Constant Acceleration Assume you started from rest but traveled further every second. Time (S): 0| 1 | 2 | 3 | 4 | 5 | S (m): 0| 0.5| 1.5 | 2.5 | 3.5 | 4.5 | t X (m) aV Vi a 1 0.5 0.5/1 = 0.5 2 2/2=1 3 4.5 4.5/3= 1.5 4 8 8/4=2 5 12.5 12.5/5 = 2.5 Calculate aV -Average velocity Instantaneous velocity Acceleration

Constant Acceleration Time (S): 0| 1 | 2 | 3 | 4 | 5 | S (m): 0| 0.5| 1.5 | 2.5 | 3.5 | 4.5 | t s av.V Vi a 1 0.5 2 2.0 1.0 3 4.5 1.5 4 8.0 5 12.5 2.5 Draw graphs of S vs. t Vi vs. t A vs. t

Constant Acceleration Time (S): 0| 1 | 2 | 3 | 4 | 5 | S (m): 0| 0.5| 1.5 | 2.5 | 3.5 | 4.5 | t s av.V Vi a 1 0.5 2 2.0 1.0 3 4.5 1.5 4 8.0 5 12.5 2.5 Draw graphs of S vs. t Vi vs. t a vs. t

Constant Acceleration Time (S): 0| 1 | 2 | 3 | 4 | 5 | x (m): 0| 0.5| 1.5 | 2.5 | 3.5 | 4.5 | t x av.V Vi a 1 0.5 2 2.0 1.0 3 4.5 1.5 4 8.0 5 12.5 2.5 Draw graphs of x vs. t Vi vs. t A vs. t

Constant Acceleration Plot the graphs by copying the table in to a spreadsheet>> ∆x (m) v (m/s) a (m/s2) 9_ 8_ 7_ 6_ 5_ 4_ 3_ 2_ 1_ x | | | | | 1 2 3 4 5 | | | | | 1 2 3 4 5 | | | | | 1 2 3 4 5 t(s) t(s) t(s)

Constant Acceleration CalDisplacement vs time Displacement increases by an increasing amount each second ∆x (m) Now calculate the gradient of this graph at a particular time t (s)

Constant Acceleration Velocity vs time graph Velocity increases by the same amount every second. Calculate the gradient of this graph and the area under it. v (m/s) +1 m/s 1s +1 m/s 1s t (s)

Constant Acceleration Acceleration vs time The graph is a horizontal straight line as the acceleration does not change. Calculate the area under this graph. a (m/s2)) t (s)

Constant Acceleration Calculate: gradient of a tangent to the displacement time graph gradient of the velocity time graph area under acceleration time graph area under velocity time graph

Constant Acceleration 2: (x2, y2) (5.6, 14) 2: (x2, y2) (6;6) 2  2  Area = l x b = 5 x 1 = 5 m/s = change in velocity! 1  1  3.5 1: (x1, y1) (0;0) 1: (x1, y1) (1.9, 0) Gradient at (3.5s) = y/  x = (14-0)/(5.6-1.9) = 3.6 m/s = instantaneous velocity at 3.5 s Gradient = y/  x = 6-0/(6-0) = 1.0 m/s2 = acceleration

Constant Acceleration Area = l x b = 5 x 1 = 5 m/s = velocity! Displacement(m) Velocity (m./s) Time (s) Time (s) Time (s) Area (0-5s)= ½ b.h = ½ (5)*(5) (s x m.s-1 = m) = 12.5 m = displacement (0-5s) Area under velocity-time graph = displacement

SKETCH GRAPHS - Constant ACCELERATION 1 2 .... . . . . . . . . . . S1 S2 time/(s) 0 1 2 3 4 5 6 7 8 9 10 Displacement increases by (uniformly ) increasing amounts in equal time intervals. ACCELERATION - TIME DISPLACEMENT - TIME VELOCITY - TIME v (m/s) a (m/s2) s2 s/(m) GRAD = … GRAD = …. s1 t1 t t/(s) t2 t/(s) t/(s) The gradient of the displacement vs time graph (tangent - at any point) gives the ……………..of the object at that point. The gradient of the velocity vs time graph (at any point) gives the ……………….of the object.

SKETCH GRAPHS - Constant ACCELERATION 1 2 .... . . . . . . . . . . S1 S2 time/(s) 0 1 2 3 4 5 6 7 8 9 10 Displacement increases by (uniformly ) increasing amounts in equal time intervals. ACCELERATION - TIME DISPLACEMENT - TIME VELOCITY - TIME v (m/s) a (m/s2) s2 s/(m) GRAD = v GRAD = a s1 t1 t t/(s) t2 t/(s) t/(s) The gradient of the displacement vs time graph (tangent - at any point) gives the velocity of the object at that point. The gradient of the velocity vs time graph (at any point) gives the acceleration of the object.

SKETCH GRAPHS - Constant ACCELERATION 1 2 -- - - - - - - - - --> Displacement increases by (uniformly ) increasing amounts in equal time intervals. DISPLACEMENT - TIME VELOCITY - TIME ACCELERATION - TIME s/(m) v (m/s) 2. a (m/s2) 2. 1. 2. 1/2bxh AREA = … AREA = …(av) l x h 1. s s2 1. s1 l x h t2 t2 t1 t2 t/(s) t1 t/(s) t1 t/(s) The AREA of the acceleration vs time graph (between any 2 points) gives the …………… ……………… of the object between the two points. The AREA under the VELOCITY vs time graph (between any 2 points) gives the ………………… of the object between the points.

SKETCH GRAPHS - Constant ACCELERATION 1 2 -- - - - - - - - - --> Displacement increases by (uniformly ) increasing amounts in equal time intervals. DISPLACEMENT - TIME VELOCITY - TIME ACCELERATION - TIME x/(m) v (m/s) 2. a (m/s2) 2. 1. 2. 1/2bxh AREA = x AREA = v (av) l x h 1. x x2 1. x1 l x h t2 t2 t1 t2 t/(s) t1 t/(s) t1 t/(s) The AREA of the acceleration vs time graph (between any 2 points) gives the change in velocity of the object between the two points. The AREA under the VELOCITY vs time graph (between any 2 points) gives the displacement of the object between the points.

SKETCH GRAPHS - Summary 1 2 -- - - - - - - - - --> DISPLACEMENT - TIME VELOCITY - TIME ACCELERATION - TIME v (m/s) a (m/s2) 2. s/(m) 1. 2. 1. t1 t2 t/(s) t1 t/(s) t2 t/(s) Gradient of the displacement vs time graph gives the …………….. AREA under the VELOCITY vs time graph gives the ……………………. …………….. of the acceleration vs time graph gives the average velocity. ……………… of the velocity vs time graph gives the acceleration of the object.

SKETCH GRAPHS - Summary 1 2 -- - - - - - - - - --> DISPLACEMENT - TIME VELOCITY - TIME ACCELERATION - TIME v (m/s) a (m/s2) 2. x/(m) GRAD = v 1. 2. GRAD = a 1. AREA = xs AREA = v t1 t2 t/(s) t1 t/(s) t2 t/(s) Gradient of the displacement vs time graph gives the velocity. AREA under the VELOCITY vs time graph gives the displacement. AREA of the acceleration vs time graph gives the change in velocity. Gradient of the velocity vs time graph gives the acceleration of the object.

Graphs of Motion A - acceleration (+) B - constant velocity (+) C - deceleration (+) D - stationary E - reverse acceleration (-) F - reverse constant v (-) G - reverse deceleration (-) ... . . . . . . . s . . . . . . . . t . . . . . . . ... v . t . . . . . .. a . . . . . . . . . t ... . . . . . . .

Graphs of Motion A - acceleration (+) B - constant velocity (+) C - deceleration (+) D - stationary E - reverse acceleration (-) F - reverse constant v (-) G - reverse deceleration (-) D ... . . . . . . . C E s B F . . . . . . . . A G t . . . . . . . ... v . t . . . . . .. a . . . . . . . . . t ... . . . . . . .

SKETCH GRAPHS - Displacement vs Time ........................................ . . . . . . . . Constant Velocity Constant Acceleration x/(m) s/(m) t/(s) t/(s) ……………………………… ………………….displacement per second - ……………… increases.. . . . . . .... ............................ Stationery Object Deceleration. Object moves backwards. s (m) s (m) s (m) t/(s) t/(s) t/(s) …………………………….. displacement ………………. - slowing down. Displacement is being…………………(…….. …………………………………….). ………………….. ……………………

SKETCH GRAPHS - Displacement vs Time ........................................ . . . . . . . . Constant Velocity Constant Acceleration x/(m) x /(m) t/(s) t/(s) Displacement increases by same amount every second. Increasing displacement per second - velocity increases. . . . . . .... x (m) t/(s) Object moves backwards. ............................ Stationery Object Deceleration. x (m) x (m) t/(s) t/(s) Rate of increase in displacement decreases - slowing down. Displacement is being reduced (coming back to start). Displacement remains constant.

SKETCH GRAPHS - Velocity vs Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v/(m) v/(m) t/(s) t/(s) . . . . . . .. . . . . . . . .. . . Constant deceleration. v (m) v (m) v (m) t/(s) t/(s) t/(s) Object ………………. ………………………. v = 0 …………………………

SKETCH GRAPHS - Velocity vs Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v/(m) v/(m) t/(s) t/(s) . . . . . . .. . . . . . . . .. . . Constant deceleration. v (m) v (m) v (m) t/(s) t/(s) t/(s) Object ………………. ………………………. v = 0 …………………………

SKETCH GRAPHS - Velocity vs Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CONSTANT VELOCITY CONSTANT ACCELERATION. v/(m) v/(m) t/(s) t/(s) . . . . . . .. . . . . . . . .. Object at REST. Constant deceleration. v (m) Acceleration backwards v (m) v (m) t/(s) t/(s) t/(s) Object accelerates backwards. v = 0 Velocity is decreasing.

SKETCH GRAPHS - Task Draw sketch graphs of displacement vs time, velocity vs time and acceleration vs time for a car that starts off from rest, accelerates uniformly for 5s, attains a speed of 30m/s which it maintains for 10s. The car then slows down for 15s at which point it comes to a halt. Indicate as much detail on the graphs as possible. x (m) a/(m/s2) v/(m/s) 5s t/(s) t/(s) t/(s)

SKETCH GRAPHS - Task Draw sketch graphs of displacement vs time, velocity vs time and acceleration vs time for a car that starts off from rest, accelerates uniformly for 5s, attains a speed of 30m/s which it maintains for 10s. The car then slows down for 15s at which point it comes to a halt. s/(m) a = v/t = (30 – 0)/5 = 6 m.s-2 v/(m/s) 30 5 15 30 t/(s)

SKETCH GRAPHS - Task Draw sketch graphs of displacement vs time, velocity vs time and acceleration vs time for a car that starts off from rest, accelerates uniformly for 5s, attains a speed of 30m/s which it maintains for 10s. The car then slows down for 15s at which point it comes to a halt. s/(m) a = v/t = (30 – 0)/5 = 6 m.s-2 v/(m/s) a/(m/s2) 6 30 5 15 30 -2 t/(s) 5 15 30 5 15 30 t/(s) t/(s)

SKETCH GRAPHS - Task Draw sketch graphs of displacement vs time, velocity vs time and acceleration vs time for a car that starts off from rest, accelerates uniformly for 5s, attains a speed of 30m/s which it maintains for 10s. The car then slows down for 15s at which point it comes to a halt. s/(m) a = v/t = (30 – 0)/5 = 6 m.s-2 v/(m/s) a/(m/s2) 6 30 5 15 30 -2 t/(s) 5 15 30 5 15 30 t/(s) t/(s) a = v/t = (0 – 30)/15 = -2 m.s-2

SKETCH GRAPHS – Example hwk Describe in words the motion of the object during each of the intervals; a, b, c, d, e and f V (m/s) a f b t (s) c e d

Describe in words the motion during each interval 8 V | 0 10 20 30 40 -12

Tut on Graphs & equations constant velocity constant velocity 8 slowing down (decelerating uniformly) accelerating forwards V | constant velocity 0 10 20 30 40 speeding up reverse uniform acceleration negative direction slowing (decelerating) down backwards -12

Graphs of two motions. The graph shows the speeds of two cars A & B as a function of time. What is the difference in the distances traveled by A & B after 20 s? (8) After how many seconds will A & B have traveled the same distance? (6) Calculate the acceleration of A. (4) V (m/s) A 6 5 B 10 15 20 t (s)

Graphs of two motions. V (m/s) 6 A 5 B t t (s) 15 20 The graph shows the speeds of two cars A & B as a function of time. What is the difference in the distances traveled by A & B after 20 s? (7) After how many seconds will A & B have traveled the same distance? (6) Calculate the acceleration of A. V (m/s) 6 A 5 B 1. DISTANCE = AREA under graph For B: area = l x b = 20 x 5 = 100 m For A: area = (1/2bh) + (l x b) = ½(15)(6) + (5 x 6) = 75 m Difference in distance travelled = 100 – 75 = 25m t t (s) 15 20 Acceleration = gradient Gradient = rise/run = (6-0)/(15-0) = 0.4 m.s-2 Distance = area at 15 s X(A) =1/2bh = ½*15*6=45m X(B) = lxb=15x5=75m X(A) = X(B) 45+t*6 = 75 +t*5 6t – 5t = 75 - 45 t = 30 s Total time = 15 +t = 15 +30 = 45s

Graphs of Vertical Motion An object is thrown vertically up and returns to the thrower’s hand. 10 20 30 -10 -20 -30 V (m/s) -10 -20 -30 30 20 10 X a V t t t