This page contains the AQA GCSE Physics P9 Motion Questions and kerboodle answers for revision and understanding .This page also contains the link to the notes and video for the revision of this topic.
9.1 Speed and Distance Time Graphs AQA GCSE Physics P9 Motion Kerboodle Answers : Page No. 135
1 a) i. For an object travelling at constant speed the distance is the length travelled by an object from a particular point. S.I Unit =
It is a scalar quantity.
ll. the gradient of its distance-time graph is basically the slope of a graph. It is calculated as the change in one quantity with respect to another quantity.
For a distance-time graph
Gradient =, which is equal to speed.
b. i. Distance
travelled by X in 1000 seconds = 30,000m
Therefore, speed = distance / time
= 30,000/1000 = 30m/s.
ii. Z
stopped for = 1000-500 = 500 sec.
iii. average speed of Z for the 1500s journey = total distance covered / total time taken
=
20,000 / 1500 = 40 / 3 = 13.33 m/s
2. A
vehicle on a motorway travels 1800 m in 60s.
a. the average speed of the vehicle in m = 1800
/ 60 = 30 m/s
b. Distance
travelled = speed * time = 30 * 300 =
9000 m
c. it
would take110 sec to travel a distance of 3300m at this speed, Time =
distance / speed = 3300 / 30 = 110 sec
3. A car
on a motorway travels a certain distance in six minutes at a speed of 21 m/s .A coach takes seven minutes to travel
the same distance. Distance and the speed
of the coach = 21 *
360 = 7560 m, speed of the coach
= distance / time = 7560 / 420 = 18m/s
4. a. A train takes 2 hours and 40 minutes to
travel a distance of 360 km.
The average speed of the train in meters per
second on this journey, Time = 160 * 60 = 9600 s
distance travelled = 360 * 1000 = 360000 m
average speed of the
trains = 360000 / 9600 = 37.5 m/s
b. The train travelled at a constant speed of 40m/s for a distance of 180km. time taken = 180000 / 40 = 4500 s, in minutes = 4500s/60 = 75 minutes
9.2 Velocity & Acceleration AQA GCSE Physics P9 Motion Kerboodle Answers : Page No. 137
1 a. By Comparing speed and velocity, Speed is a scalar quantity i.e. it only has magnitude and do not have any direction whereas velocity is a vector quantity it have both direction and magnitude.
b. A car
on a motorway is travelling at a constant speed of 30m/s when it overtakes a lorry travelling at a speed of 22
m/s. If both vehicles maintain their
speeds,speed of car = 30m/s, distance travelled by car in 300 sec
= 300 * 30 = 9000 m, distance travelled by lorry in 300 sec = 300 * 22 = 6600 m, distance between car and lorry = 9000 –
6600 = the car will be 2400 m ahead of
the lorry after 300s.
2. The
velocity of a car increased from 8m/s to 28m/s in 16s without change of direction. Acceleration =
change in velocity / time taken = 28 – 16 /16
= 0.75
𝑠^25 m/eleration = change in velocity / time taken = 28 – 16 /16
on and magnitude.istance time graph.gardo points.
3. The
driver of a car increased the speed of the car as it joined the motorway, it then travelled at constant velocity
before slowing down as it left the
motorway at the next junction.
a i. the car decelerated, When it
left the motorway at the next junction.
ii. the
acceleration of the car was zero, When it
was on the motorway.
b. When
the car joined the motorway, it accelerated from a speed of 7.0m/s for 10s at
an acceleration of 2.0m/s’.
speed at the end of this time is calculated as follows,
a = v – u/t
therefore v = u + at
v = 7 + 2 *
10 = 27 m/s
4. A
sprinter in a 100m race accelerated from rest and reached a speed of 9.2m/s in the first 3.1 s.
a. the
acceleration of the sprinter in this time, a = 9.2 – 0 / 3.1 = 2.96
b. The
sprinter continued to accelerate to top speed and completed the race in 10.4s. . Average speed = 100 / 10.4 = 9.61 m/s.
9.3 More about Velocity-time Graphs AQA GCSE Physics P9 Motion Kerboodle Answers : Page No. 139
After
matching each of the following descriptions (i to iv) to one of the lines,
labeled A, B, C and D on the velocity – time graph (Figure 3).
i. Accelerated motion throughout. – B
ii. Zero acceleration. – A
iii. Accelerated motion, then decelerated
motion. – D
iv. Deceleration throughout. – C
2. Look
at Figure 3.
a.
identify the line representing the object that travelled:
i. the furthest distance – A
ii. the least distance. – C
b. B
travelled further.
3. a. Distance travelled by object A = area under the line A
i.e. area of the rectangle = 8*20 =
160m.
b. Distance travelled by object B = 1/2*20*8 = 80m.
4. a.Distance
travelled by object C = 1/2*2*20 = 20m.
b. Distance
travelled by D = 2*1/2*6*10 = 60m, Distance
travelled by A =160m. Difference between
A & D = 160 – 60m = 100m.
9.4 Analyzing Motion Graphs AQA GCSE Physics P9 Motion Kerboodle Answers : Page No. 141
1 a .
Distance in 8 sec = 120m,
the speed of the object shown on the graph in
Figure 1= 120/8
= 15m/s.
b. the
speed of the object shown in Figure 2 is continuously increasing.
2. The
graph in Figure 4 shows how the velocity of a cyclist on a straight road changes with time.
a. The speed of the cyclist
is constant upto 40 sec and then the cyclist decelerates upto 60 sec and velocity becomes zero.
b. i. the
acceleration of the cyclist in the first
40s. Acceleration = 8-0 /40 = 0.2
ii. the acceleration of the cyclist in the following 20s. acceleration = 0 – 8 /20 = -0.4
3.In a
motorcycle test, the speed from rest was recorded at intervals in Table 1.
a. velocity-time
graph
b. the
initial acceleration of the motorcycle
= 10 – 0 / 5 = 2
c. i. Movement of motorcycle in first 20 seconds. 10 * 40 = 400 m
ii. Movement
of motorcycle in the following 10 seconds= 10 * 40 = 400m
4
u = 0 ; s = 1000m ; a = 2
therefore v =
velocity = 63.24
Summary Questions:- Page no. 142
1. A model car travels around a circular track at constant speed. If given the equipment, we can measure the speed of car. The boundary of the track will be measured by tape which is the distance covered by the car. We can mark a point on the track by marker. We measure the time by the stopwatch when the car crosses the mark first time till it crosses the point consequently. By distance and time figures we can calculate the speed of the car.
2. a. Part A to B of the journey was faster
b I Distance travelled between A & B = 2000m ; time = 100sec. = 20m/s
ii. Distance travelled between B & C = 500m ; time = 50sec.
The speed of the car between B & C= 500/50 = 10m/s.
c. If the car had travelled the whole distance of 2500m at the same speed as it travelled between A & B. Distance travelled = 2500m ; speed = 20m/s.
Thus, time taken = 2500/20 = , the journey would have taken 125 seconds.
3. Figure 2 shows the distance- time graph for a motorcycle approaching a speed limit sign.
a. Firstly upto 5 seconds, the speed increases and consequently it decelerated.
b. In one
second, the distance travelled = 20m. So, the speed = 20/1 =
20m/s.
ii. Distance travelled in 10 sec. = 120m. Thus, speed = 120/ 10 = 12m/s.
4 a. A car took 10s to increase
its velocity from 5m/s to 30m/s. a= v-u /
t acceleration of the car is = 30 –
5 / 10 = 2.5
b. i . Distance
travelled = 1/2 *10*25 + 5*10
=
125+50 = the car travelled 175m.
ii. the
average speed of the car in this time
= 35-5/10 = 2.5 m/s.
5. a. velocity-tome graph
b. the train’s acceleration in each of the three parts of the journey is calculated as follows:
In the first part, From 0 to 80 sec. = 20/80 = 0.25
2nd part, From 80 to 120 sec. the speed is constant, thus acceleration = 0.
Third part, From 120 to 160 sec. = -20/40 = – 0.50
c. the total distance travelled by the train = 1/2*20*80 + 20*40 + 1/2*20*40 = 2000m.
d. the average speed for the train’s journey was 12.5 m/s. We can show this by the formula = Total distance / Total time = 2000 / 160 = 12.5m/s.
6. A water skier started from rest and accelerated steeply to 12 m/s in 15 sec. , then travelled at constant speed for 45sec. before slowing down steeply and coming to a halt 90s after she started.
a. The velocity -time graph for this journey.
Ans.
b. the acceleration of water skier in first 15 seconds= 12 – 0 / 15 = 0.8
c. the deceleration of the water skier in the final 30 seconds= 0 -12 / 30 = -0.4
d. Total distance travelled by the skier = 1/2*15*12 + 12*45 + 1/2*12*30
= 810 m/s.
Practice Questions:- Page no. 143
1.1 total distance travelled by the truck.= 600m.
1.2 the total time taken to offload all the parcels = 300seconds.
1.3 Between points D & E truck travelling at the greatest speed because slope or gradient of the graph between D& E is highest.
1.4 Between C & E.the truck change speeds without stopping.
1.5 the average speed of the truck over the complete journey = Total distance covered / total time taken,= 600m/ 300s. = 2m/s.
1.6 Advantage of using an electric truck rather than a diesel truck in a warehouse. An electric truck does not release harmful gases thus prevents pollution and suffocation in the warehouse.
2.1 the difference between the terms speed and velocity, Speed is a scalar quantity whereas velocity is a vector quantity.
2.2 A firework rocket is sent up into the air, the velocity-time graph of the journey is shown. The average acceleration of the rocket between 0 – 40 seconds = 30 – 0 / 40
= 0.75
2.3 1/2* 40*30 + 30*20 + 1/2*60*30 = 2100m.
2.4 The rocket is accelerating between points A & B.
The rocket is stationary between points B & C.
The rocket is decelerating between points C &D.
3. Engineers are testing a new water slide in an aqua park. They are using a plastic dummy and measuring the time it takes for the dummy to travel from top to bottom. The rate of running water through the slide is being changed.
3.1 Length of the water slide i.e. the distance is independent variable.
3.2 Time taken by the dummy to slide the distance is dependent variable.
3.3 This is not a suitable test because, the changing rate of water does not provide accurate result of the time taken.
3.4 In one test the dummy is sliding at 10 m/s at the end of the slide. The total slide length is 20m.The acceleration of the dummy through the slide. v= 10m/s ; u = 0m/s ; s = 20m.
a = = 100 – 0 / 2*20 = 100/40
= 2.5
4. Physical Quantity
1. Speed – Scalar
2. Mass – Scalar
3. Acceleration – Vector
4. Weight – Scalar
5. Energy – Scalar
5. Competitors are taking part in a triathlon race which involves running, swimming and cycling in immediate succession over a 50km distance. There are many factors that may affect the eventual outcome of the race. The average running speed is 8km/hr. The average cycling speed is 15.5 km/hr, The average swimming speed is 5.92 km/hr. The factors which may affect the performance of all competitors are the weather conditions, the wind velocity in same direction or opposite direction, the speed and direction of water current.
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References:
BBC Bitesize
AQA GCSE Science Kerboodle textbook
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