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Answer: 2.5 m s⁻² (shown); t = 2.5 sคำตอบ: 2.5 m s⁻² (shown); t = 2.5 s
1(a) Size of the acceleration1(a) ขนาดของความเร่ง
Unit vector $\mathbf{i}$ points due east and $\mathbf{j}$ due north. Positions are measured from the fixed point $O$.
A particle $P$ has constant acceleration. At $t=0$ it is at $O$, moving with velocity $(2\mathbf{i}-3\mathbf{j})$ m s⁻¹. At $t=2$ s it is at the point $A$, whose position vector is $(7\mathbf{i}-10\mathbf{j})$ m.
(a) Show that the acceleration of $P$ has size 2.5 m s⁻². [4]
1(b) Time to head north-east1(b) เวลาที่วิ่งไปทางตะวันออกเฉียงเหนือ
Unit vector $\mathbf{i}$ points due east and $\mathbf{j}$ due north. Positions are measured from the fixed point $O$.
A particle $P$ has constant acceleration. At $t=0$ it is at $O$, moving with velocity $(2\mathbf{i}-3\mathbf{j})$ m s⁻¹. At $t=2$ s it is at the point $A$, whose position vector is $(7\mathbf{i}-10\mathbf{j})$ m.
(b) At $A$ the acceleration of $P$ changes to the constant value $(4\mathbf{i}+8.8\mathbf{j})$ m s⁻². Later, at the point $B$, $P$ is heading north-east. Find the time $P$ takes to go from $A$ to $B$. [4]
Answer: 2√5 ≈ 4.47 m s⁻¹; T = 10; c = 10คำตอบ: 2√5 ≈ 4.47 m s⁻¹; T = 10; c = 10
2(a) Speed at B2(a) อัตราเร็วที่ B
Here $\mathbf{i}$ and $\mathbf{j}$ are unit vectors in the horizontal plane, and positions are measured from a fixed origin $O$.
A particle $P$ slides on a smooth horizontal plane with constant acceleration $(2.4\mathbf{i}+\mathbf{j})$ m s⁻². At $t=0$ it is at the point $A$, position vector $(44\mathbf{i}-10\mathbf{j})$ m, with velocity $(-16\mathbf{i}-3\mathbf{j})$ m s⁻¹. At $t=5$ s it is at the point $B$.
(a) Find the speed of $P$ at $B$. [4]
2(b) Find T from the i parts2(b) หา T จากส่วน i
Here $\mathbf{i}$ and $\mathbf{j}$ are unit vectors in the horizontal plane, and positions are measured from a fixed origin $O$.
A particle $P$ slides on a smooth horizontal plane with constant acceleration $(2.4\mathbf{i}+\mathbf{j})$ m s⁻². At $t=0$ it is at the point $A$, position vector $(44\mathbf{i}-10\mathbf{j})$ m, with velocity $(-16\mathbf{i}-3\mathbf{j})$ m s⁻¹. At $t=5$ s it is at the point $B$.
(b) At time $t=T$ s, with $T\gt 5$, $P$ is at the point $C$ with position vector $(4\mathbf{i}+c\mathbf{j})$ m, where $c$ is a constant. Find the value of $T$. [3]
2(c) Find c from the j parts2(c) หา c จากส่วน j
Here $\mathbf{i}$ and $\mathbf{j}$ are unit vectors in the horizontal plane, and positions are measured from a fixed origin $O$.
A particle $P$ slides on a smooth horizontal plane with constant acceleration $(2.4\mathbf{i}+\mathbf{j})$ m s⁻². At $t=0$ it is at the point $A$, position vector $(44\mathbf{i}-10\mathbf{j})$ m, with velocity $(-16\mathbf{i}-3\mathbf{j})$ m s⁻¹. At $t=5$ s it is at the point $B$.
(c) $P$ is at the point $C$, with position vector $(4\mathbf{i}+c\mathbf{j})$ m, at time $t=10$ s (the value of $T$ from part (b)). Find the value of $c$. [3]
Answer: F = i − 1.2j N; v = −3i + 3j m s⁻¹; 315°; OA = −10i + 9j mคำตอบ: F = i − 1.2j N; v = −3i + 3j m s⁻¹; 315°; OA = −10i + 9j m
3(a)(b) Force, then velocity3(a)(b) แรง แล้วต่อด้วยความเร็ว
Unit vector $\mathbf{i}$ points due east and $\mathbf{j}$ due north.
A particle $P$ of mass 0.5 kg moves on a smooth horizontal plane. Its acceleration is the constant $(2\mathbf{i}-2.4\mathbf{j})$ m s⁻², and the only horizontal force on it is a constant force $\mathbf{F}$ N.
(a)(b) Find $\mathbf{F}$ in terms of $\mathbf{i}$ and $\mathbf{j}$ (1 mark). At $t=0$, $P$ has velocity $(-7\mathbf{i}+7.8\mathbf{j})$ m s⁻¹. Then find the velocity of $P$ at $t=2$ s (2 marks). [3]
3(c) Direction as a bearing3(c) ทิศเป็นแบริ่ง
Unit vector $\mathbf{i}$ points due east and $\mathbf{j}$ due north.
A particle $P$ of mass 0.5 kg moves on a smooth horizontal plane. Its acceleration is the constant $(2\mathbf{i}-2.4\mathbf{j})$ m s⁻², and the only horizontal force on it is a constant force $\mathbf{F}$ N.
(c) At $t=2$ s the velocity of $P$ is $(-3\mathbf{i}+3\mathbf{j})$ m s⁻¹. Find the direction of motion of $P$ at that time, as a bearing in degrees. [3]
3(d) Position vector OA3(d) เวกเตอร์ตำแหน่ง OA
Unit vector $\mathbf{i}$ points due east and $\mathbf{j}$ due north.
A particle $P$ of mass 0.5 kg moves on a smooth horizontal plane. Its acceleration is the constant $(2\mathbf{i}-2.4\mathbf{j})$ m s⁻², and the only horizontal force on it is a constant force $\mathbf{F}$ N.
(d) At $t=0$, $P$ is at the fixed point $O$, moving with velocity $(-7\mathbf{i}+7.8\mathbf{j})$ m s⁻¹, and at $t=5$ s it is at the point $A$. Find $\overrightarrow{OA}$ in terms of $\mathbf{i}$ and $\mathbf{j}$. [2]
Answer: 6√6 ≈ 14.7 m s⁻¹; a = 6t i − 3t^(−1/2) j m s⁻²; −62i + 24j mคำตอบ: 6√6 ≈ 14.7 m s⁻¹; a = 6t i − 3t^(−1/2) j m s⁻²; −62i + 24j m
4(a)(b) Speed, then acceleration4(a)(b) อัตราเร็ว แล้วความเร่ง
Positions are measured from a fixed origin $O$.
At time $t$ seconds, $t\gt 0$, a particle $P$ has velocity $\mathbf{v}=(3t^2\mathbf{i}-6t^{\frac{1}{2}}\mathbf{j})$ m s⁻¹. At $t=4$ the position vector of $P$ is $(\mathbf{i}-4\mathbf{j})$ m.
(a)(b) Find the speed of $P$ when $t=2$ (2 marks). Then find the acceleration of $P$ at time $t$, in terms of $t$, $\mathbf{i}$ and $\mathbf{j}$ (2 marks). [4]
4(c) Position at t = 14(c) ตำแหน่งเมื่อ t = 1
Positions are measured from a fixed origin $O$.
At time $t$ seconds, $t\gt 0$, a particle $P$ has velocity $\mathbf{v}=(3t^2\mathbf{i}-6t^{\frac{1}{2}}\mathbf{j})$ m s⁻¹. At $t=4$ the position vector of $P$ is $(\mathbf{i}-4\mathbf{j})$ m.
(c) Work out where $P$ is at $t=1$, as a position vector. [4]
Answer: a = (3/2)t^(−1/2) i − 2j m s⁻²; t = 9/4 s; r = (2t^(3/2) − 2) i − t² j m; 2√113 m (= √452)คำตอบ: a = (3/2)t^(−1/2) i − 2j m s⁻²; t = 9/4 s; r = (2t^(3/2) − 2) i − t² j m; 2√113 m (= √452)
5(a)(b) Acceleration, then direction5(a)(b) ความเร่ง แล้วทิศทาง
Positions are measured from a fixed origin $O$.
At time $t$ seconds, $t\gt 0$, a particle $P$ has velocity $\mathbf{v}=(3t^{\frac{1}{2}}\mathbf{i}-2t\mathbf{j})$ m s⁻¹.
(a)(b) Find the acceleration of $P$ at time $t$ (2 marks). Then find the value of $t$ at the instant when $P$ is moving in the direction of $\mathbf{i}-\mathbf{j}$ (3 marks). [5]
5(c) Position r in terms of t5(c) ตำแหน่ง r ในรูป t
Positions are measured from a fixed origin $O$.
At time $t$ seconds, $t\gt 0$, a particle $P$ has velocity $\mathbf{v}=(3t^{\frac{1}{2}}\mathbf{i}-2t\mathbf{j})$ m s⁻¹.
(c) When $t=1$, the position vector of $P$ is $-\mathbf{j}$ m. Find $\mathbf{r}$, the position vector of $P$ at time $t$, in terms of $t$. [3]
5(d) Distance when speed is 105(d) ระยะเมื่ออัตราเร็วเป็น 10
Positions are measured from a fixed origin $O$.
At time $t$ seconds, $t\gt 0$, a particle $P$ has velocity $\mathbf{v}=(3t^{\frac{1}{2}}\mathbf{i}-2t\mathbf{j})$ m s⁻¹.
(d) Using $\mathbf{r}=\left((2t^{\frac{3}{2}}-2)\mathbf{i}-t^2\mathbf{j}\right)$ m from part (c), find, in exact form, how far $P$ is from $O$ at the instant its speed is 10 m s⁻¹. [6]
Answer: c = 3 (shown); 3√17/4 ≈ 3.09 m s⁻¹; T = 9คำตอบ: c = 3 (shown); 3√17/4 ≈ 3.09 m s⁻¹; T = 9
6(a) Show c = 36(a) แสดงว่า c = 3
Show every stage of your working: an answer that rests only on a calculator earns no marks.
Unit vector $\mathbf{i}$ points due east and $\mathbf{j}$ due north, and positions are measured from a fixed origin $O$.
For $t\ge 1$, with $t$ in seconds, the position vector of a particle $P$ is $\mathbf{r}=\left(ct^{\frac{1}{2}}\mathbf{i}-\dfrac{3}{8}t^2\mathbf{j}\right)$ m, where $c$ is a constant. When $t=4$, $P$ is on a bearing of $135^\circ$ from $O$.
(a) Show that $c=3$. [3]
6(b) Speed at t = 46(b) อัตราเร็วเมื่อ t = 4
Show every stage of your working: an answer that rests only on a calculator earns no marks.
Unit vector $\mathbf{i}$ points due east and $\mathbf{j}$ due north, and positions are measured from a fixed origin $O$.
For $t\ge 1$, with $t$ in seconds, the position vector of a particle $P$ is $\mathbf{r}=\left(ct^{\frac{1}{2}}\mathbf{i}-\dfrac{3}{8}t^2\mathbf{j}\right)$ m, where $c$ is a constant. When $t=4$, $P$ is on a bearing of $135^\circ$ from $O$.
(b) Using $c=3$, find the speed of $P$ when $t=4$. [4]
6(c) Find T from the direction6(c) หา T จากทิศของความเร่ง
Show every stage of your working: an answer that rests only on a calculator earns no marks.
Unit vector $\mathbf{i}$ points due east and $\mathbf{j}$ due north, and positions are measured from a fixed origin $O$.
For $t\ge 1$, with $t$ in seconds, the position vector of a particle $P$ is $\mathbf{r}=\left(ct^{\frac{1}{2}}\mathbf{i}-\dfrac{3}{8}t^2\mathbf{j}\right)$ m, where $c$ is a constant. When $t=4$, $P$ is on a bearing of $135^\circ$ from $O$.
(c) With $c=3$, at $t=T$ the acceleration of $P$ points in the direction $(-\mathbf{i}-27\mathbf{j})$. Find the value of $T$. [4]
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