1. 2.2 Equations of uniformly accelerated motion
Skid marks
Uniformly accelerated motion
Derivation of equations of motion
Check-point 5 1 2
Book 2 Section Equations of uniformly accelerated motion

2. Book 2 Section 2.2 Equations of uniformly accelerated motion
Skid marks
In an accident, drivers brake their cars hard, skid marks are left on the road.
The police will measure the length of the skid marks. Why?
To estimate the speed of car before it brakes.
Book 2 Section Equations of uniformly accelerated motion

3. 1 Uniformly accelerated motion
Consider an object accelerating uniformly from initial velocity u to final velocity v over time t.
Book 2 Section Equations of uniformly accelerated motion

4. 1 Uniformly accelerated motion
Average velocity (v ) =
u + v 2
Displacement
= area under v-t graph (i.e. a trapezium) =
(u + v)t 2
Book 2 Section Equations of uniformly accelerated motion

5. 2 Derivation of equation of motion
Consider a general case:
uniform acceleration = a
initial velocity = u
final velocity = v
time = t
Four important equations for uniformly accelerated motion can be derived.
Book 2 Section Equations of uniformly accelerated motion

6. 2 Derivation of equation of motion
By the definition of acceleration,
a =
v – u t
……(i)
Re-arrange (i),
v = u + at
……(1)
Area under the graph equals displacement s, 1 2
s =
 (u + v)  t
……(2)
Book 2 Section Equations of uniformly accelerated motion

7. 2 Derivation of equation of motion
Sub (1) into (2),
s =
ut + 1 2
at 2
…(3)
v – u
Sub
t =
into (2), a 1 2
v – u a
s =
 (u + v) 
2as = (v + u)  (v – u)
v 2 = u 2 + 2as
…(4)
Book 2 Section Equations of uniformly accelerated motion

8. 2 Derivation of equation of motion
The four equations obtained are:
v = u + at 1 2
s =
(u + v)  t
ut +
at 2
v 2 = u 2 + 2as
They are called the equations of motion.
Book 2 Section Equations of uniformly accelerated motion

9. 2 Derivation of equation of motion
When applying the four equations,
1 acceleration must be constant;
2 the signs of s, u, v and a should be consistent with the defined +ve direction.
Example 9
Velocity of a powerboat
Book 2 Section Equations of uniformly accelerated motion

10. 2 Derivation of equation of motion
Example 10
Road test on brake
Book 2 Section Equations of uniformly accelerated motion

11. 2 Derivation of equation of motion
Stopping distance, thinking distance and braking distance
Example 11
Book 2 Section Equations of uniformly accelerated motion

12. Book 2 Section 2.2 Equations of uniformly accelerated motion
Check-point 5 – Q1
True or false:
A car accelerates from t1 to t2 non-uniformly from v1 to v2 over distance d.
The average velocity of the car is
v1 + v2 2 .
(T / F)
Book 2 Section Equations of uniformly accelerated motion

13. Book 2 Section 2.2 Equations of uniformly accelerated motion
Check-point 5 – Q2
Maria (1.5 m tall) is standing under an apple tree. An apple 2.5 m right above her falls down with an acceleration of 10 m s–2.
Find the time the apple takes to hit her.
Book 2 Section Equations of uniformly accelerated motion

14. Book 2 Section 2.2 Equations of uniformly accelerated motion
Check-point 5 – Q2 ?
u = ________
t = ________
10 m s–2
v = ________ ?
a = ________
2.5 – 1.5 = 1 m
s = ______________
v = u + at 1 2
s =
(u + v)  t
ut +
at 2
v 2 = u 2 + 2as 
1 =
0 + 1 2
(10)t 2
t =
t = s
Book 2 Section Equations of uniformly accelerated motion

15. Book 2 Section 2.2 Equations of uniformly accelerated motion
Check-point 5 – Q3
(a) Compared with stopping on a dry road, a car stopping on a wet road will have a longer ___________ distance and ___________ distance.
braking
stopping
(b) If the deceleration of the car is halved, the stopping distance (will/will not) be doubled.
Book 2 Section Equations of uniformly accelerated motion

16. Book 2 Section 2.2 Equations of uniformly accelerated motion
Check-point 5 – Q4
Blackbird (Lockheed SR-71) was the fastest jet in the world.
Max. speed = 894 m s–1
Max. acceleration
= 3.75 m s–2
Estimate the shortest distance travelled when it accelerates from rest to its top speed.
Book 2 Section Equations of uniformly accelerated motion

17. Book 2 Section 2.2 Equations of uniformly accelerated motion
Check-point 5 – Q4 ?
u = ________
t = ________
v = ________
894 m s–1
a = ________
3.75 m s–2 ?
s = ________
v = u + at 1 2
s =
(u + v)  t
ut +
at 2
v 2 = u 2 + 2as 
8942 = 0 + 2(3.75)s
s = m
Book 2 Section Equations of uniformly accelerated motion

18. Book 2 Section 2.2 Equations of uniformly accelerated motion
The End
Book 2 Section Equations of uniformly accelerated motion