1 Straight-line motion and graphs

Skills check 3. a) s

1
1. (20 + 60) × 30 = 1200 11
2
30 − 15 3
2. = = 0.375
40 8
5
10 − 30 1
= – = –0.5
40 2

30 8 O
3. a) = 6s b) = 40 s 10 20 30 t
5 0.2
5 b) Total distance = 11 m
c) = 0.2 s
25 4. The particle travels 15 m in 10 s at constant
4. a) 12 × 12 = 144 m b) 0.4 × 10 = 4 m velocity, then it is stationary for 30 s before it
returns to its start point at constant velocity in 30 s.
c) 30 × 0.5 = 15 m
15
24 a) v = = 1.5 m s–1
45 10
5. a) = 3 m s–1 b) = 1.5 m s–1
8 30
b) v = 0
10
c) = 0.2 m s–1 15
50 c) v = – = –0.5 m s–1
30
6. a) 10 × 10 = 100 m s–1 b) 0.2 × 30 = 6 m s–1
5. The train travels 8 m for 5 s at constant velocity,
5 32 − 44 then it is stationary for 15 s and then travels 12 m
7. a) = 3 m s–2 b) = –2 m s–2
5 6 in 15 s at constant velocity.
60 − 15 8
8. = 3s a) v = = 1.6 m s–1
15 5
b) v = 0
Exercise 1.1 c) v =
12
= 0.8 m s–1
15
1. a) s
20 6. The elevator descends 9 m in 8 s at constant
velocity, then is stationary for 8 s then it descends
9 m in 12 s at constant velocity.
9
a) v = – = –1.125 m s–1
8
O 10 20 30 t
b) v = 0
−20
b) v = = –4 m s–1 9
5 c) v = – = – 0.75 m s–1
12
2. a) s 7. a) s = 40 × 8 = 320 m

200
320
b) v = – = –16 m s–1
20
20
8. a)= 30
T
2
O
10 15 23 t T = = 0.667
3
−200
b) v = = –25 m s–1
8

© Oxford University Press 2018: this may be reproduced for class use solely for the purchaser’s institute Straight-line motion and graphs 1
 b) It takes 3T seconds to complete the journey. 1
b) s = (130 + 50) × 5 = 450 m
2
T = 0.667
450
c) v = = 3.46 m s–1

3T = 2, so time taken to complete journey is 2 s. 130
16
S S 5. a) a = = 1.6 m s−2 at t = 5 s
9. a) =4 ∴T= 10
4T 16
24
−S S 8
a= = 1.6 m s−2 at t = 30 s
v = = − S ÷ = −S × = − 8m s−1 15
2T 8 S
1
27 b) s = × 10 × 16 + 10 × 16 = 80 + 160 = 240 m
9 2
b) T = = so 4T = 18 1
6 2 c) s = 240 + (16 + 40) × 15
S 2
Therefore 4 = s = 240 + 420
18
so S = 18 × 4 = 72 s = 660 m
16 1
10. a) V = = 1.6 6. a) × 60 × V = 450; V = 15 m s−1
10 2
15

b) For the next 5 seconds, the train travels back b) a = = 0.75 m s−2
20
at 1.6 m s–1, therefore it covers 5 × 1.6 = 8 m 1
and the displacement from A to B is –8 m. c) When t = 40 v = V = 7.5
2
1 1
c) Between C and D the train travels at s = × 20 × 15 + (15 + 7.5) × 20
2 2
1.5V = 2.4 m s–1 s = 150 + 225
Displacement from C to D = 5 × 2.4 = 12 m s = 375 m
d) At D the displacement is 16 – 8 + 12 = 20 m. v 6
From D to E, displacement = –20 m. 7. a) t = = = 30 s
a 0.2
20 1
e) Velocity from D to E = – = –2 m s–1 b) s = (180 + 70) × 6 = 750 m
10 2
750
c) v = = 4.17 m s−1
Exercise 1.2 180
v 5
1. 8. a) a1 = = 2.5 m s−2 a2 = −2.5 m s−2
30
2
a3 = −2.5 m s−2 a4 = 2.5 m s–2
1
b) s = (7 + 3) × 5 = 25 m
2
1
O c) s = (9 + 5) × 5 = 35 m
20 30 t 2

2. v d) The first stop is after 7 seconds when it has

20 gone 25 m and therefore travelled 10 floors
2 4
upwards. It is now on the 18th floor.
O t It stays here until moving downwards 35 m
between t = 10 and t = 19. This means it moves
–20 35

down = 14 floors, so it finished on the
2.5
3. v
4th floor.
24
1
(10 + 25)V = 420
9. a)
2
35
V = 420
O
10 70 82 t 2
V = 24 m s–1
5 1 24
4. a) a = = = 0.667 m s–2 b) a = = 3.43 m s–2
30 6 7
5 −24
a = – = –0.1 m s–2 c) a = = –3 m s–2
50 8

© Oxford University Press 2018: this may be reproduced for class use solely for the purchaser’s institute Straight-line motion and graphs 2
 v b) When V = 4
d) t =
a
6 T = 8 − 4 = 4 s
t1 = = 1.75 s
3.43
6 13. a) Greatest height is when v = 0. So when t = 2.
t2 = 25 – = 25 – 2 = 23 s 1
3 s = × 2 × 20 = 20 m
2
1 1
10. a) (30 + V  ) × 10 + 10V + × 16V = 380 Height is 20 + 25 = 45 m above the ground.
2 2
150 + 5V + 10V + 8V = 380 1
(T – 2)V1 = – 45
b)
2 1
23V = 230
V1
V = 10 m s−1 = –10, V1 = –10(T1 – 2)
−20 T1 − 2
b) When t = 8 a = = −2.0 m s–2
10 1
−10
– (T – 2) × 10(T1 – 2) = – 45
When t = 30 a = = −0.625 m s–2 2 1
16
T12 – 4T1 + 4 = 9
380
c) v = = 10.6 m s–1
36 T12 – 4T1 – 5 = 0
15 −V (T1 – 5) (T1 + 1) = 0
11. a) = 2.5
2T
T1 = 5 s (T1 is positive)
⎛ V⎞
T = ⎜ 3 − ⎟ c) V1 = –10 (5 – 2) = –30 m s–1
⎝ 5⎠
1 1 1
(15 + V  )2T + VT + (V + 15)T = 70 d) V2 = – V1 = 15 m s–1
2 2 2
(15 + V  )3T + 2VT = 140 e) Greatest height is when v = 0.
(45 + 5V  )T = 140 The time taken for the velocity to reach
(45 + 5V  ) ⎛⎜ 3 − ⎞⎟ = 140
V 15
0 is = 1.5 s.
⎝ 5⎠ 10
135 + 15V – 9V – V 2 = 140 1
So the greatest height is s = × 15 × 1.5
V – 6V + 5 = 0
2 2
(V – 5)(V – 1) = 0 = 11.25 m
V = 5 m s–1 or V = 1 m s–1 f) To fall back from 11.25 m
1
b) When V = 5 11.25 = 2 × 10t2

5
T = 3 – = 2 s
t = 1.5
5
The total time to come to rest is
When V = 1
5 + 1.5 + 1.5 = 8 s
1
T = 3 – = 2.8 s
5 Summary exercise 1
s
12. a) 8 − V = 1 ⇒ T = 8 − V 1. a) = 10
T 0.04
s = 0.4 m
Distance = area under graph = area of
0.4
rectangle + 2 triangles b) =8
t
1 1
25V + (8 − V )T + (10 − V )10 = 138 t = 0.05 s
2 2
1 1
s
25V + (8 − V ) + (10 − V )10 = 138
2
0.4
2 2
50V + V 2 − 16V + 64 + 100 − 10V = 276
V 2 + 24V − 112 = 0
(V + 28)(V − 4 ) = 0 O
0.04 0.09 t
−1
Since V > 0, V = 4 m s

© Oxford University Press 2018: this may be reproduced for class use solely for the purchaser’s institute Straight-line motion and graphs 3
2. a) s 1
800 × 12 × t = 20
b)
2
S 10
t = = 3.33 s
3
12
0 a= = –3.6 m s–2, so the
5 15 T tB tA t
−3.33
deceleration is 3.6 m s–2
b) The train from B changes speed when
t = 5 + 10 = 15 s 5. a) v
40

When t = 15 the position of B is 800 – 10 × 20
= 600 m from A. V

When the two trains meet, when t = T, the O 10 20 36 t

position of the train from B is
40 −V
S = 600 – 25(T – 15) b) =2
10
The position of the train from A is S = 20T 40 – V = 20, V = 20 m s–1
Solving 600 − 25 (T − 15) = 20T
c) The total distance travelled
600 − 25T + 375 = 20T 1 1
975 = 45T = 
(40 + 20) × 10 + 20 × 10 + × 16 × 20
2 2
T  21.7 s = 300 + 200 + 160
Hence S= 20 × 21.7 = 660 m
S  433 m from A 660 1
The average velocity is = 18 = 18.3 m s–1
36 3
c) The train from A will reach B when
6. a) v
t = 800 ÷ 20 = 40 s
20
The train from B will reach A when
t = 15 + 600 ÷ 25 = 39 s 12

Hence the train from B passes through the V

station at the opposite end of the track first.
O T 15 25 t
3. a) The distance of the descent is
1 b) In the f  inal phase of motion
× 8 × 30 = 120 m
2
20 − V
1 = 1.8
b) (10 + 26) V = 120 25 − 15
2
18V = 120 20 – V = 1.8 × 10
V = 6.67 m s –1 V = 2.0 m s–1
30 c) From the area under the graph
c) a = = 15 m s–2
2 1 1
1 (12 + 2)T + 2(15 −T ) + (2 + 20)10 = 179
d) s = × 6 × 30 = 90 m 2 2
2 14 220
T + 2(15 −T ) + = 179
1 1 2 2
4. a) × 1.5 × 8 + (8 + 12) × 3
2 2 7T + 30 − 2T +110 = 179
+ 12 (T – 4.5) = 100 5T = 39
6 + 30 + 12T – 54 = 100 T = 7.8
12T = 118
T = 9.83 s

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