Speeding up, slowing down?

Speeding up, slowing down

Speeding up, slowing down

Physics on the RoadLesson 7

LI…Describe uniform acceleration.Use kinematic equations to

make predictionsInterpret motion graphs

Constant accelerationWhen something gains speed at a constant

rate this is called uniform acceleration. Lets begin with motion in a straight line

vv∆vv+∆v∆v

∆t∆t∆t t=0

vv +∆v

v +2∆v

Traffic Incident

A policeman using a video camera and radar claims that a car starting from some lights reached 90 km/h when it passed a shop 150m away 12 s later. The driver claims this is impossible.

Distance =150 mInitial velocity = 0 m/sFinal velocity = 90 km/h

= 25 m/sTime = 12 s

Check 1: many cars advertise a 0-100 km/h of less than 10s

Check 2: The car would need to gain 25/12=2.08m/s each of the 12 seconds or 2.08 m/s/s

metres per second per second is the unit of acceleration and is abbreviated to ms-2.

Traffic Incident cont…A policeman using a video camera and radar claims that a car starting from some lights reached 90 km/h when it passed a shop 150m away 12 s later. The driver claims this is impossible.

Distance =150 mInitial velocity = 0 m/sFinal velocity = 25 m/sTime = 12 s

Car accelerating to 25m/s in 12s

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9 10 11 12

Time /s

speed /

m/s

∆v

∆t

acceleration = change of velocity time

acceleration = 25 = 2.08 ms-2

12

avg. speed = change of distance time

avg. speed = 150 = 12.5 ms-1

12

avg. speed = 0 + 25 = 12.5 ms-1

2

distance = avg. speed x time = 12.5 x 12= 150m

Linking to distance time graphs Draw this velocity

time graph Use your graph and

the relationship between average speed and distance travel to complete a distance time table.

Plot the distance time graph

Car accelerating to 25m/s in 12s

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9 10 11 12

Time /s

speed /

m/s

distance = avg. speed x time = 12.5 x 12= 150m

Time /s Distance /m

12 150

Distance travelled is the areaunder a velocity time graph

Kinematic equations

acceleration = change of velocity time

acceleration = final velocity – initial velocity time

a = v – ut

at= v – u

v= u + at

Car accelerating to 25m/s in 12s

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9 10 11 12

Time /s

speed /

m/s

Distance travelled is the area under the graph

Area of triangle = ½ x base x height = ½ x time x change in velocity

= ½ x 12 x 25 = 150 m

First equationWhat if the object wasn’t originally stationary?

More kinematic equations

velocity time graph

012345678

0 1 2 3 4

time (s)

velo

city

(m

/s)

The distance travelled is the area under the graph

Area A + Area B

A

B

u

v

u x t + ½ x t x (v-u)

but… at = v – uso…

s = ut + ½ at2

2nd

equation

Third equation of motion

v= u + att= v – u a

average velocity= u + v 2

Distance = average velocity x timeSo….

s=u+v t 2s = (u+v )x (v-u) 2 as = v2 – u2

2a

(u+v)(v-u)= v2 – u2

v2 = u2 + 2as3rd equation

1st equation

Newton’s equations of motion

v= u + at v2 = u2 + 2ass = ut + ½ at2

Use to computefinal velocity from acceleration and time.

Use to computedistance from acceleration and time.

Use to computeFinal velocity from acceleration and distance.

v – final velocity (ms-1) t – time (s)u – initial velocity (ms-1) s –distance/displacement (m) a – acceleration (ms-2)

Questions

v= u + at

v2 = u2 + 2as

s = ut + ½ at2

A car travelling at 10 ms-1 accelerates at 3ms-2 for 5 seconds what is the car’s final velocity?.

How far will the car travel whilst accelerating?

A coin is dropped from the Centre Point tower in Sydney. The tower is 309m high. Acceleration due to gravity is 10 ms-2. What speed does the coin hit the ground? What assumption have you made?

Q 1 a Q 1 b

Q 2