Velocity

1. Block 2
Distance, Speed and
Acceleration

2. Displacement, Velocity
and Acceleration
Special Rates of Change

3. What is to be learned?
• How rates of change apply to
displacement, velocity and time.

4. Displacement
Distance (in a certain direction)
Velocity
Change in displacement over a period of
time
Acceleration
Change in velocity over a period of time
x
v = dx
/dt
a = dv
/dt
(some call it speed!)
Original Formula
Derivative of Displacement
Derivative of Velocity

5. Ex An object is moving along x axis (cm) at
time t(secs) according to equation:
x = 4t2
+ 10t –t3
Calculate:
a) Displacement, Velocity and
Acceleration
after 1 second and 3 seconds
b) Displacement and Velocity after
5 seconds

6. x = 4t2
+ 10t –t3
displacement
t = 1
x = 4(1)2
+ 10(1) – 13
= 13cm
velocity
v = dx
/dt = 8t + 10 – 3t2
t = 1
v = 8(1) + 10 – 3(1)2
= 15 cm/sec
Original Formula
Derivative of Displacement

7. x = 4t2
+ 10t –t3
velocity
v = dx
/dt = 8t + 10 – 3t2
t = 1 v = 8(1) + 10 – 3(1)2
= 15 cm/sec
acceleration
a = dv
/dt = 8 – 6t
t = 1 a = 8 – 6(1)
= 2 cm/sec/sec
cm/sec2
Derivative of Velocity

8. x = 4t2
+ 10t –t3
displacement
t = 3 x = 4(3)2
+ 10(3) – 33
= 39cm
velocity
v = dx
/dt = 8t + 10 – 3t2
t = 3 v = 8(3) + 10 – 3(3)2
= 7 cm/sec

9. x = 4t2
+ 10t –t3
velocity
v = dx
/dt = 8t + 10 – 3t2
t = 3 v = 8(3) + 10 – 3(3)2
= 7 cm/sec
acceleration
a = dv
/dt = 8 – 6t
t = 3 a = 8 – 6(3)
= -10 cm/sec2

10. x = 4t2
+ 10t –t3
velocity
v = dx
/dt = 8t + 10 – 3t2
t = 3 v = 8(3) + 10 – 3(3)2
= 7 cm/sec
acceleration
a = dv
/dt = 8 – 6t
t = 3 a = 8 – 6(3)
= -10 cm/sec2
It is decelerating

11. x = 4t2
+ 10t –t3
displacement
t = 5 x = 4(5)2
+ 10(5) – 53
= 25cm
velocity
v = dx
/dt = 8t + 10 – 3t2
t = 5 v = 8(5) + 10 – 3(5)2
= -25 cm/sec

12. x = 4t2
+ 10t –t3
displacement
t = 5 x = 4(5)2
+ 10(5) – 53
= 25cm
velocity
v = dx
/dt = 8t + 10 – 3t2
t = 5 v = 8(5) + 10 – 3(5)2
= -25 cm/sec
It has changed direction

13. Displacement
Distance (in a certain direction)
Velocity
Change in displacement over a period of
time
Acceleration
Change in velocity over a period of time
x (or h)
v = dx
/dt (0r dh
/dt)
a = dv
/
Motion and Derivatives
(some call it speed!)
Original Formula
Derivative of Displacement
Derivative of Velocity

14. Ex Belinda throws a ball into the air
Its height (h m) after t secs is:
h = 4t – t2
Find
a) Its height, velocity and
acceleration after 1 sec
b) What is its height when the
velocity is zero?

15. h = 4t – t2
height (displacement)
t = 1 h = 4(1) – 12
= 3m
velocity
v = dh
/dt = 4 – 2t
t = 1 v = 4 – 2(1)
= 2m/sec

16. h = 4t – t2
velocity
v = dh
/dt = 4 – 2t
t = 1 v = 4 – 2(1)
= 2m/sec
acceleration
a = dv
/dt = -2
t = 1 (or whatever!) a = -2 m/sec2
decelerating by 2m/sec2

17. b) Velocity zero?
v = dh
/dt = 4 – 2t
 4 – 2t = 0
4 = 2t
t = 2
velocity is 0 m/sec after 2 seconds
height?
h = 4t – t2
t = 2 h = 4(2) – 22
= 4 m

18. Graphical explanation
h = 4t – t2
t
h
1
3
v = 0
4
2

19. The displacement (x cm) of a point after t secs can be
measured by the formula x = 10 – 6t + t2.
Calculate when the velocity will reach 10 cm/sec and
what the displacement will be after this time?
v = dx
/dt = -6 + 2t
→ -6 + 2t = 10
→ 2t = 16
→ t = 8 secs
t = 8, x = 10 – 6(8) + 82
= 26 cm
Tricky Key
Question