Distance and Displacement (m) displacement-time graphs SCUVAT HOW to de
displacement: shortest distance ↳ velocity = rate of ✗ = displacement •
x
from A → B in a certain direction. change of displacement equations u = initial velocity
average, total distance grad-velocity • v- ut at v = final velocity ☒ 9=-9-8
A speed time • x (Utr) t a = acceleration.
t 5ms-I
↳ scalar t: time
average = total displacement distance-time graphs • ut + Eat?
Displacement:O velocity time ↳ speed: rate of • ✓ 2=42+292 119
If the object travels ↳ vector d change of distance 29
A→ B then B. → A. grad = speed From rest- initial velocity = 0
Instantaneous Velocity - Acceleration up or down under EXPERIMENT
a velocity at a certain point in time t gravity -19.81ms-2 DUE TO GRA
✓ = rate of change of displacement ÷ time Only need 3 out of 5 scurat quantities. method 7:
calculate gradient of x-t graph Acceleration At max height: V-O - drop marble
↳ use tangent on a curve at certain time ↳ rate of change of velocity 1m52) - record time
OBJECTS ACCELERATING - measure he
1:-a
STOPPING DISTANCES using scurat a = positive, if accelerating in direction of motion -measure tim
stopping = thinking + braking Velocity - time Graph a- negative, if accelerating in opposite direction - Plot a table
distance ↳ decelerating
Thinking = distance travelled by vehicle from the
V
grad-acceleration E. g an object thrown up against gravity or h t E
moment the driver sees the hazard and press' brake area- displacement a car braking
Braking -- distance travelled from the moment of Falling Under Gravity: y-mate
the driver pressing the brake to the car stopping t up = - 9.81ms-2 h=½gt²+O
• braking point hazard distances Instantaneous acceleration down = 9.81ms-2 - Plot a graph
car T-thinking V
find gradient of - gradient = 2
I B B- braking tangent PROJECTILE MOTION lif project
car collisions: horizontal + vertical components are independent! acts on
t
IE 2+2 ax energy = KE=½mv² consider if Q uses air resistance or not. vertical
✗= B SO B - V? 42 if velocity increases resultant use trigonometry velocity
29 V
by factor of 2 FACTORS EFFECTING THINKING DISTANCE: vertical = sin(o) ✗ r Max h
v=O as car stops: B =_ 42 KE=½M(ZV12 • alcohol, tiredness, distractions (phone), drugs ⑨ horizontal = cos (o) ✗ r until it
a = negative (deceleration) 29 then KE quadruples FACTORS EFFECTING BRAKING DISTANCE: h Horizont
stopping distance = T + B SO KE = 4×12 MV2 • tyre conditions, mass of vehicle, weather - [assume no air resistance] speed
Driving twice as fast increases Sd by a factor squared road conditions if wet or slippery. to end
displacement: shortest distance ↳ velocity = rate of ✗ = displacement •
x
from A → B in a certain direction. change of displacement equations u = initial velocity
average, total distance grad-velocity • v- ut at v = final velocity ☒ 9=-9-8
A speed time • x (Utr) t a = acceleration.
t 5ms-I
↳ scalar t: time
average = total displacement distance-time graphs • ut + Eat?
Displacement:O velocity time ↳ speed: rate of • ✓ 2=42+292 119
If the object travels ↳ vector d change of distance 29
A→ B then B. → A. grad = speed From rest- initial velocity = 0
Instantaneous Velocity - Acceleration up or down under EXPERIMENT
a velocity at a certain point in time t gravity -19.81ms-2 DUE TO GRA
✓ = rate of change of displacement ÷ time Only need 3 out of 5 scurat quantities. method 7:
calculate gradient of x-t graph Acceleration At max height: V-O - drop marble
↳ use tangent on a curve at certain time ↳ rate of change of velocity 1m52) - record time
OBJECTS ACCELERATING - measure he
1:-a
STOPPING DISTANCES using scurat a = positive, if accelerating in direction of motion -measure tim
stopping = thinking + braking Velocity - time Graph a- negative, if accelerating in opposite direction - Plot a table
distance ↳ decelerating
Thinking = distance travelled by vehicle from the
V
grad-acceleration E. g an object thrown up against gravity or h t E
moment the driver sees the hazard and press' brake area- displacement a car braking
Braking -- distance travelled from the moment of Falling Under Gravity: y-mate
the driver pressing the brake to the car stopping t up = - 9.81ms-2 h=½gt²+O
• braking point hazard distances Instantaneous acceleration down = 9.81ms-2 - Plot a graph
car T-thinking V
find gradient of - gradient = 2
I B B- braking tangent PROJECTILE MOTION lif project
car collisions: horizontal + vertical components are independent! acts on
t
IE 2+2 ax energy = KE=½mv² consider if Q uses air resistance or not. vertical
✗= B SO B - V? 42 if velocity increases resultant use trigonometry velocity
29 V
by factor of 2 FACTORS EFFECTING THINKING DISTANCE: vertical = sin(o) ✗ r Max h
v=O as car stops: B =_ 42 KE=½M(ZV12 • alcohol, tiredness, distractions (phone), drugs ⑨ horizontal = cos (o) ✗ r until it
a = negative (deceleration) 29 then KE quadruples FACTORS EFFECTING BRAKING DISTANCE: h Horizont
stopping distance = T + B SO KE = 4×12 MV2 • tyre conditions, mass of vehicle, weather - [assume no air resistance] speed
Driving twice as fast increases Sd by a factor squared road conditions if wet or slippery. to end
