Therefore V2 = 2 x V av
V av = (V1 +
V2) / 2
V av = (V1 + V2)
2
V2 = 10 x 2s = 20 ms-1
iii. Acceleration.
The increase of speed is from zero to 20
ms-1.
The time taken for this was 10 s.
Acceleration= Change of velocity / Time.
Acceleration = (20 ms-1 -
0 ) / 10
s.
Acceleration = 2 ms-2
More about Graphs.
Fig. 4 Uniform velocity graphs.
Graphs
|
Displacement
/ Time gra
|
Velocity / Tim
|
Shape
|
A straight line showing an upward slope.
|
A straight line parallel to the time axis.
|
Gradient
|
Gr = d/t = 50 m / 5 s
= 10 ms1
|
Zero
|
Displacement
|
Height at any time gives the displacement.
|
The area below the graph. ( This part is coloured green to
make it clear)
|
Solution.
Fig.5 A Graph showing Acceleration.
Make your Comments on the two Graphs A and B and compare with what is given.
Graph A | Graph B |
i. Not a straight line. The shape is an upward curve.
ii. The gradient of the graph is changing. This increase is due to an increase in velocity.
| i. A straight line inclined up. ii. The area below the graph gives the distance traveled. iii. Area of green triangle. Area = ½ base x height. A=1/2 ( 5x10) =50. M2 ( Although the area unit is square meters here the answer has to be in meters. |
Formulae
Symbols:- Average velocity = V. av. First velocity V 1 ,
Second velocity= V.2 Time = t., Displacement = d, Acceleration=a
V = d / t. | V.av = (Vi + V.2) / 2 for uniform velocity. | V.f = Vi + a.t |
a = (V 2 – Vi ) / t. | d = Vit + ½ at2 | |
Multiple Choice Questions.
1.0 Which of these will give a vector quantity?
A- Distance / Time. B- Displacement /Time. C- Length X Breadth. D- Change in speed/Time.
2.0 Which is the correct definition for acceleration?
A- Velocity/ time. B- Speed / Time. C- Change in velocity / time taken. D- displacement/ Time.
The graph for the questions from 3.0. to 5.0

3.0 The graph shows,
4.0 What is the speed during the first 3 seconds?
A- 10 ms-1 , B- 90 ms -1 C- 5 ms-1 . D- 10 ms -2
5.0 What is the speed during the last 2 seconds?
A- 10 ms-1 , B- 90 ms -1 C- 5 ms-1 . D- 10 ms -2
The graph for the questions 6.0 to 9.0.
Graph for questions 6 to 9.
6.0 What is the acceleration during the first 5 seconds?
A - (45 - 20) / 5 ms-2. B- 45 /5 ms-2.
C- 0 ms-2. D- 1/2 X 5 X 25 ms-2
7.0 What is the displacement during the first 5 seconds?
A - (45 20) / 5 m. B- 45 /5 m.
C- 5 x 45 m. D- 1/2 X 5 X 25 ms-2
8.0 What is the average velocity?
A. 45 /5 ms-1 B . 45 - 20 /5 ms-1 C. 45 /2 ms-1 D. (45 + 2o) /2
9.0 What is the correct statement about he graph?
A- Shows regular velocity. B- Shows regular acceleration.
C- Shows increasing acceleration. D- Shows decreasing acceleration.
10.0 An object is dropped and it falls to the ground for 6 seconds. Taking the acceleration of gravity as 9.8 ms-2 , what is the distance it has fallen? ( Neglect air friction)
A- d = ½ 9.8 x 6x6. m. B- 6 x 9.8 / 6 m. C- 9.8 x2 x 6 m. D - 9.8./2 x 6 m .
ANSWERS
1, B, 2. C, 3. B, 4. A, 5. A 6. A. 7. D. 8. D 9. B 10.
General questioons.
Q.1.0
A train 50 m. long passes a signal post in 2 seconds. Find the following:
i. The velocity (speed) of the train.
ii. What is the time it will take to pass a platform 30 m long?
iiii. If the train comes to rest in 5 seconds the (negative) acceleration
on braking. (deceleration)
Q.2.0
A plane bypassing an aerodrome, cruising at 150 m.s-1 was observed for 5 s.
On a command by the control tower it reduced the speed (velocity) and landed in 10 seconds.
i. What would be the displacement at cruising speed?
ii. What is the negative acceleration in landing?
iii. What could be the displacement in stopping?
Q.3.0 The downward velocity of a bungee jumper has been recorded and plotted as shown.
i. What is the maximum velocity attained? ii. Find the positive acceleration .. iii. Find the acceleration during the reducing speed. iv. Find the height freely fallen before the chord tightens.
Q. 4.0
The braking system of a car is capable of a deceleration at the rate of 6 m.s-2
Find the stopping distance when traveling at 30 ms-1
Q.5.0
Examine the Velocity / Time grpha here and answer the questions below
i. Find the average velocity for each second separately.
II. Find the distance traveled .
Time | One second | Two seconds. | Three second | Four seconds |
Displacement | | | | |
iii. Draw a displacement /Time graph.
Time / s | 1 | 2 | 3 | 4 |
Displacement | | | | |
ANSWERS for general Questions
Q.1.0
i. Velocity = distance/ time
V = 50/ 2 V= 25 ms-1. | ii.Time = 50 + 30 /25 = 80/25 = 3.2 s. | iii. a = Change in velocity / time = 25 /5 = - 5 m.s-2 |
i. Velocity = distance/ time
V = 50/ 2 V= 25 ms-1. | ii.Time = 50 + 30 /25 = 80/25 = 3.2 s. | iii. a = Change in velocity / time = 25 /5 = - 5 m.s-2 |
Q.2.0
i.
V = d / t. 150 = d x 5 ∴ d = 150 x 5 = 750 m. | ii. a =(V f – Vi ) / t. a =(0 - 150) / 10 = -15 m.s-2 | iii. method 1. d = V1t + ½ at2 d = (150 x 10) + -15x100 d= 1500 -1500/2 = = 1500 - 750 = 750 m Method 2. V.av = (Vi + V.f) / 2 = 75 m/s d= V xt = 750 m.
|
Q.3.0
i. What is the maximum velocity attained? ii. Find the positive acceleration .. iii. Find the acceleration during the reducing speed. iv. Find the height freely fallen before the chord tightens.
i. 40 ms-1 | ii. a =(V 2 – Vi ) / t. = (40 - 0) / 4.5 = 8.9 ms-2 | iii. a =(V 2 – Vi ) / t = (0- 40) /4.5 = - 8.9 ms-2
| iv. d =V.av x t V.av = 1/2 (V f +Vi ) = 1/2 (40 + 0) = 20 ms-1
d =V.av x t = 20 x 4.5 =4.4 m.
|
Q. 4.0
The braking system of a car is capable of a deceleration at the rate of 6 m.s-2
Find the stopping distance when traveling at 30 ms-1
Finding the time. a =(V f – Vi ) / t. 6 = (0 -30) /t 6t = -30 t= -5 s. | Finding the distance. d = V av x t. d = 15 x 5 = 75
|
5.0
i. Find the average velocity for each second separately.
Period | 1 | 2 | 3 | 4 |
Velocity avg. m/s | 5/2 =2.5 | 15/2 7.5 | 25/2 12.5 | 35/2 17.5 |
II. Find the distance traveled .
Time | One second | Two seconds. | Three second | Four seconds |
Displacement | 2.5 m. | 10 m. | 22.5 | 40 m |
iii. Draw a displacement /Time graph