graphs describing - · pdf filegraphs describing motion pg 31, 34 ... the gradient of the...

Graphs describing motion

Pg 31, 34 - 36

© cgrahamphysics.com 2015

Sketching and interpreting motion graphs

Topic 2: Mechanics 2.1 – Motion

EXAMPLE: Suppose Freddie the Fly begins at x = 0 m, and travels at a constant velocity for 6 seconds as shown. Find two points, sketch a displacement vs. time graph, and then find and interpret the slope and the area of your graph. SOLUTION: •The two points are (0 s, 0 m) and (6 s, 18 m). •The sketch is on the next slide.

x = 0 x/m x = 18 t = 0, t = 6 s,




SOLUTION: •The slope is rise over run or 18 m / 6 s •Thus the slope is 3 m s-1, which is interpreted as Freddie’s velocity.

0 3 6 9

12 15 18 21 24 27

x / m

t / s 0 1 2 3 4 5 6 7 8 9

s = 18 - 0 s = 18 m

Rise

Run t = 6 - 0 t = 6 s


An object moving with constant speed

0

2

4

6

8

10

12

1 2 3 4 5 6 7 8 9 10 11

dis

tan

ce (

m)

time(s)

D- t graph

stopped

Negative slope à coming back

The graph gives average speed, which is equal to the instantaneous speed at all points


Non uniform motion

�  If the velocity is not constant, the d-t graph is a curve

�  To find the instantaneous speed at 3s, we need to find the gradient of the tangent at 3s

�  Taking the tangent means that we let ∆𝑣= ∆𝑠/∆𝑡  𝑎𝑠 𝑡→0, which is differentiation

0 2 4 6 8

10 12 14 16 18

1 2 3 4 5 6

distan

ce (m

)

-me (s)

D-‐t graph

Gradient = 10−2/4.7−2 =2.96

The blue line represents the average velocity

If the graph is in the form s = 𝑘𝑡↑2 , 𝑑𝑠/𝑑𝑡 =2𝑘𝑡 à gradient at any time is equal to instantaneous speed © cgrahamphysics.com 2015

Summary

�  The instantaneous velocity at a point on a positive position time graph is the slope of the tangent drawn at that point

�  The average velocity between any two points on a d – t graph for any kind of motion is equal to the line joining the two points 𝑣↓𝑎𝑣 = 16.5 −2/6 −1 =2.9𝑚𝑠↑−1 

d-t v – t a - t

slope slope

Area under graph © cgrahamphysics.com 2015

Velocity – time graph

�  𝑎= ∆𝑣/∆𝑡 = 𝑑𝑣/𝑑𝑡  �  𝑎= 𝑑/𝑑𝑡 (𝑑𝑠/𝑑𝑡 )= 𝑑↑2 𝑠/𝑑𝑡↑2  

Changing speed or velocity =

acceleration


Constant speed / velocity

�  v= constant �  a = 0




EXAMPLE: Calculate and interpret the area under the given v vs. t graph. Find and interpret the slope.

SOLUTION: •The area of a triangle is A = (1/2)bh. •Thus

A = 1/2 ×20 ×30 = 300 m. •This is the displacement of the object in 20 s. •The slope is (30 m/s) / 20 s = 1.5 m s-2. •This is the acceleration of the object.

VE

LOC

ITY

(ms-

1 )

0

50 40 30 20 10

t TIME (sec)

0 5 10 15 20


Example

�  A train accelerates uniformly from rest to reach a speed of 45m𝑠↑−1  in a time of 3.0 min. It then travels at this speed a further 4.0 min at which time the brakes are applied. It comes to rest with constant acceleration in a further 2.0 min

�  Draw a v-t graph for the journey

0 5

10 15 20 25 30 35 40 45 50

0 200 400 600 V

(m/

s)

T(s)

Train journey


�  From the graph calculate a) magnitude of acceleration between 0 and 3.0min

�  a=gradient = 45−0/180−0 =0.25𝑚𝑠↑−2 

�  b) magnitude of acceleration after the brakes are applied

�  a=gradient = 0−45/540−420 =−0.38𝑚𝑠↑−2 

�  c) total distance traveled by the train

�  s=area under the graph = 1/2 ×180×45+240×45+ 1/2 ×120×45 =4050+10800+2700=17550m


Acceleration time graph

�  For an object in free fall

0

5

10

15

0 2 4 6

a(m

/s2

) t(s)

Free fall

Speed/velocity is the area under the graph v=at


Example

�  The acceleration of an object increases uniformly at a rate of 3.0 𝑚𝑠↑−2 every second. If the object starts at rest calculate its speed after 10s

Solution �  Speed=area under graph

= 1/2 ×10×25 = 125 𝑚𝑠↑−2 

0 5

10 15 20 25 30 35 40 45

0 10 20

a 𝑚𝑠^

(−2)

T(s)

acceleration


Motion of a dropped ball O=dropped à uniform acceleration due to “g” A=hit ground à large acceleration from + to – B= velocity is zero when acceleration changes direction BC: less velocity than in AB CD: ball accelerates with ”g” D: maximum height, v=0, a=g

•  v at E is decreasing because of air resistance •  OA // CE //FG // HJ à gradient of these lines = g •  AC //EF //GH à gradient =acceleration of ball

whilst in contact with surface

No perpendicular line à means a=∞

a@A,C,E ≠𝑔 © cgrahamphysics.com 2015

graphs describing - · pdf filegraphs describing motion pg 31, 34 ... the gradient of the...

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