distance, speed and time - city harvest church distance, speed and time question 1: the graph shows...
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Revision Distance, Speed and Time
Question 1:
The graph shows the journey of a train from Punggol Station to Orchard Station.
a) How long did the train service break down?
b) How far is the train from Orchard Station after 5 minutes?
c) How much time would the train need to reach Orchard Station if it has not broken
down?
d) A bus is also traveling from Punggol Station to Orchard Station. It left Punggol
station at 0910 and traveled at a constant speed of 40 km/h. Show the journey of
the bus on the above graph.
Answers:
a) 0935 β 0910 = 25 ππππ b) 15 km
c) 20 ππππ d) π·
π=20
40=1
2
Revision Distance, Speed and Time
Question 2:
The diagram shows the distance-time graph for Alvinβs journey and Benβs journey from
X to Y along a walking track.
a) What is the distance between X and Y?
b) Estimate the distance between Alvin and Ben when t = 15.
c) Find Alvinβs speed when t = 12.
d) Find Alvinβs speed when t = 25.
Answers:
a) 1000m
b) 600 β 250 = 350π
c) Zero
d) 1000β600
30β15=400
15= 26
2
3π/πππ
Revision Distance, Speed and Time
Question 3:
The diagram shows the speed-time graph of a moving object.
a) Find the acceleration of the object when t = 3.
b) Given that the object decelerated at 2 m/s2. Find the value of T.
c) Calculate the average speed for the whole journey.
Answers:
a) gradient =24β4
5β0=20
5= 4π/π 2
b) gradient = 10β0
24βπ= β2
10 β 0
24 β π= β2
β5 = 24 β π
π = 29 seconds
c) Total Distance = 418
d) Speed = 418
29= 14.4π/π
Revision Distance, Speed and Time
Question 4:
A particle travels at a constant speed of 30 m/s for 10 seconds and was then brought to
rest with a uniform acceleration after a further 10 seconds.
a) On the axes provided below,
sketch the speed-time graph
of the particle for the first
20 seconds.
b) Calculate the acceleration in the last 10 seconds.
c) Calculate the speed of the particle when t = 13.7 s.
d) On the axes provided below, sketch the distance-time graph of the same journey.
Answers:
b) 30β0
10β20=
30
β10= β3π/π 2
c) ?β0
13.7β20= β3
?= 18.9π/π
d)
Revision Distance, Speed and Time
Question 5:
The diagram shows the speed-time graph of a car. The car starts from rest and attain a
speed of 40 m/s in 14 seconds. It then moves with constant speed for 20 seconds and
then slows down at a steady rate to stop at T seconds. The total distance travelled by the
car is 1480 m.
a) Find the value of T.
b) Find the deceleration of the car.
c) Find the distance travelled by the car in the first 20 seconds.
d) Draw a distance-time graph for the same journey.
Answers:
a) 1480 = (1
2Γ 14 Γ 40) + (20 Γ 40) + (
1
2Γ 40 Γ π)
1480 = 280 + 800 + 20π
20 = π
π = 34 + 20 = 54
b) 40β0
34β54=
40
β20= β2π/π 2
β΄ 2π/π 2
c) (1
2Γ 14 Γ 40) + (6 Γ 40)
= 520
d)
Revision Distance, Speed and Time
Question 6:
The diagram below shows the speed-time graph of a car, which starts from rest.
a) Calculate the speed of the car after 9 seconds.
b) Find the value of u, if the total distance covered in the first 40 seconds is 900 m.
Answers:
a) 20β0
15β0=?β0
9β0
20
15=?
9
12 =?
b) (1
2Γ 15 Γ 20) + (25 Γ 20) + π = 900
π = 250
1
2Γ 25 Γ βπ‘ = 250
βπ‘ = 20 β΄ π’ = 40
Revision Distance, Speed and Time
Question 7:
The diagram below shows a velocity-time graph
of a moving object over a period of t seconds.
a) Calculate
(i) the retardation during the first 2 seconds,
(ii) the value of t if the total distance moved for the t seconds is 36 m.
b) On the axes in the answer space, sketch the distance-time graph of the object over
the period of the t seconds.
Answers:
a) (i)
9β3
0β2=
6
β2= β3π/π 2
β΄ 3π/π 2
(ii) (1
2Γ 2 Γ 6) + (3 Γ 2) + (3 Γ 3) + π = 36
π = 15 1
2Γ π Γ 3 = 15
π = 10 π‘ = 15
b)
Revision Distance, Speed and Time
Question 8:
The diagram below shows the speed-time graph of a car which starts from rest.
a) Calculate the speed of the car at t = 9 s.
b) Find the value of v, if the total distance covered in the first 40 seconds is 900 m.
c) Find the retardation at t = 50 s
d) Use the diagram provided below to draw the distance-time graph of the car for the
whole journey.
Answers:
a)
20β0
15β0=?β0
9β0β
20
15=?
9
β΄?= 12π/π
b) (1
2Γ 20 Γ 15) + (25 Γ 20) + (
1
2Γ 25 Γ βπ‘) = 900
650 + 12.5βπ‘ = 900 βπ‘ = 20
β΄ π£ =40π
π
c) 40β0
40β60=β40
20= β
2π
π 2
β΄ 2π/π 2
d)
Revision Distance, Speed and Time
Question 9:
The speed, v m/s, of a car at time t seconds is given by v = 2t + 3 for .
For , the speed of the car decreases uniformly until it finally comes to rest.
The speed time graph is given below.
(a) Find the maximum speed of the car in km/h.
(b) Calculate the speed of the car during the 8th second.
Answers:
70 t
157 t
a) Max speed when π‘ = 7 π£ = 14 + 3 = 17π/π
π£ =17
1000Γ 3600 = 61.2ππ/β
b) Gradient =17β0
7β15=17
β8= β2
1
8π/π
?β0
8 β 15=17
β8
?= 147
8π/π
Revision Distance, Speed and Time
Question 10:
The diagram is the speed-time graph of an object during a period of 16 seconds.
a) Calculate the speed of the object when t = 4.
b) Find the distance of the whole journey.
Answers:
a) gradient =14β8
4β6= β1π/π 2
?β8
4 β 6= β1
?= 10π/π
b) (1
2Γ 6 Γ 6) + (6 Γ 8) + (6 Γ 8) + (
1
2Γ 4 Γ 8)
= 130π
Revision Distance, Speed and Time
Question 11:
The diagram is the speed-time graph of a van's journey for a period of 60 seconds.
a) If the van accelerates at 0.88 m/s2 at t = 20, find the value of v.
b) Calculate the average speed of the van.
c) In the axes given in the answer space, sketch the distance-time graph of the van for
the entire journey.
Answers:
Speed (m/s)
Time, t (s) 15 40 60
20
v
a) π£β20
40β15= 0.88
π£ = 42π/π b) Total distance = 300 + 775 + 420
Average Speed =1495
60
= 2411
20π/π
c)
Revision Distance, Speed and Time
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