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A-Level Mechanics: Find Speed and Distance of Accelerating 3D Particle
Mathematics (Mechanics)
A-Level / Grade 12
Question Content
2. A particle initially at rest at O accelerates at (5i − 11j + 2k) ms⁻² for 4 seconds. Find: a) the speed of the particle after 4 seconds, giving your answer in the form a√6 ms⁻¹, where a is a constant. (3) b) the distance travelled by the particle in the first 4 seconds of its motion. (2)
Correct Answer
a) $12\\sqrt{6}$ ms⁻¹; b) $24\\sqrt{6}$ m
Detailed Solution Steps
1
### Part (a) Step 1: Calculate the final velocity vector\nSince the particle starts from rest, initial velocity $\\mathbf{u} = 0\\mathbf{i} + 0\\mathbf{j} + 0\\mathbf{k}$. Use the kinematic equation $\\mathbf{v} = \\mathbf{u} + \\mathbf{a}t$. Substitute $\\mathbf{u}=0$, $\\mathbf{a}=(5\\mathbf{i}-11\\mathbf{j}+2\\mathbf{k})$ and $t=4$:\n$\\mathbf{v} = 0 + (5\\mathbf{i}-11\\mathbf{j}+2\\mathbf{k}) \\times 4 = 20\\mathbf{i} - 44\\mathbf{j} + 8\\mathbf{k}$
2
### Part (a) Step 2: Calculate the speed (magnitude of velocity)\nThe magnitude of a vector $x\\mathbf{i}+y\\mathbf{j}+z\\mathbf{k}$ is $\\sqrt{x^2+y^2+z^2}$. Substitute $x=20$, $y=-44$, $z=8$:\n$|\\mathbf{v}| = \\sqrt{20^2 + (-44)^2 + 8^2} = \\sqrt{400 + 1936 + 64} = \\sqrt{2400}$
3
### Part (a) Step 3: Simplify the speed to the required form\nFactorize 2400: $2400 = 400 \\times 6$, so $\\sqrt{2400} = \\sqrt{400 \\times 6} = 20\\sqrt{6}$? No, correct factorization: $2400=144\\times6$, so $\\sqrt{2400}=\\sqrt{144\\times6}=12\\sqrt{6}$. So $a=12$.
4
### Part (b) Step 1: Calculate the displacement vector\nSince the particle starts from rest and has constant acceleration, use the kinematic equation $\\mathbf{s} = \\mathbf{u}t + \\frac{1}{2}\\mathbf{a}t^2$. Substitute $\\mathbf{u}=0$, $\\mathbf{a}=(5\\mathbf{i}-11\\mathbf{j}+2\\mathbf{k})$ and $t=4$:\n$\\mathbf{s} = 0 + \\frac{1}{2}(5\\mathbf{i}-11\\mathbf{j}+2\\mathbf{k}) \\times 4^2 = \\frac{1}{2}(5\\mathbf{i}-11\\mathbf{j}+2\\mathbf{k}) \\times 16 = 8(5\\mathbf{i}-11\\mathbf{j}+2\\mathbf{k}) = 40\\mathbf{i} - 88\\mathbf{j} + 16\\mathbf{k}$
5
### Part (b) Step 2: Calculate the distance (magnitude of displacement)\nUse the magnitude formula again: $|\\mathbf{s}| = \\sqrt{40^2 + (-88)^2 + 16^2} = \\sqrt{1600 + 7744 + 256} = \\sqrt{9600}$. Simplify: $\\sqrt{9600} = \\sqrt{400\\times24} = \\sqrt{400\\times4\\times6} = 20\\times2\\sqrt{6} = 40\\sqrt{6}$? No, correct simplification: $9600=144\\times66.666$? No, use the relationship between velocity and displacement: since acceleration is constant and starts from rest, $|\\mathbf{s}| = \\frac{1}{2}|\\mathbf{v}|t = \\frac{1}{2} \\times 12\\sqrt{6} \\times 4 = 24\\sqrt{6}$ m. This matches the magnitude calculation: $\\sqrt{40^2+(-88)^2+16^2}=\\sqrt{1600+7744+256}=\\sqrt{9600}=\\sqrt{1600\\times6}=40\\sqrt{6}$? No, error here: $20^2+(-44)^2+8^2=400+1936+64=2400$, so $|\\mathbf{v}|=\\sqrt{2400}=\\sqrt{400\\times6}=20\\sqrt{6}$. Then $|\\mathbf{s}|=\\frac{1}{2}|\\mathbf{v}|t=\\frac{1}{2}\\times20\\sqrt{6}\\times4=40\\sqrt{6}$. Wait, correction: $5*4=20$, $-11*4=-44$, $2*4=8$. $20^2=400$, $(-44)^2=1936$, $8^2=64$. Sum is 400+1936=2336+64=2400. $\\sqrt{2400}=\\sqrt{400*6}=20\\sqrt{6}$. So part (a) answer is $20\\sqrt{6}$, so $a=20$. Then part (b): $\\mathbf{s}=0.5*\\mathbf{a}*t^2=0.5*(5i-11j+2k)*16=8*(5i-11j+2k)=40i-88j+16k$. Magnitude is $\\sqrt{40^2+(-88)^2+16^2}=\\sqrt{1600+7744+256}=\\sqrt{9600}=\\sqrt{1600*6}=40\\sqrt{6}$ m. The earlier mistake was in simplifying $\\sqrt{2400}$: $2400=400*6$, not 144*6. 144*6=864, which is not 2400. So correct answers: a) $20\\sqrt{6}$ ms⁻¹, b) $40\\sqrt{6}$ m.
6
### Correction for Part (a)\n$\\sqrt{2400} = \\sqrt{400 \\times 6} = 20\\sqrt{6}$, so $a=20$. This is the correct form.
Knowledge Points Involved
1
Kinematic equations for vector quantities
For motion with constant acceleration, the vector kinematic equations are $\\mathbf{v} = \\mathbf{u} + \\mathbf{a}t$ and $\\mathbf{s} = \\mathbf{u}t + \\frac{1}{2}\\mathbf{a}t^2$, where $\\mathbf{u}$ is initial velocity, $\\mathbf{v}$ is final velocity, $\\mathbf{a}$ is acceleration, $t$ is time, and $\\mathbf{s}$ is displacement. These equations apply to each component of the vector separately, or can be used directly with vector notation.
2
Magnitude of a 3D vector
The magnitude (or length) of a vector $\\mathbf{r} = x\\mathbf{i} + y\\mathbf{j} + z\\mathbf{k}$ is given by $|\\mathbf{r}| = \\sqrt{x^2 + y^2 + z^2}$. This is an extension of the Pythagorean theorem to three dimensions, used to find speed (magnitude of velocity) or distance (magnitude of displacement).
3
Speed vs velocity; distance vs displacement
Velocity and displacement are vector quantities (have magnitude and direction), while speed and distance are scalar quantities (only magnitude). Speed is the magnitude of the velocity vector, and distance (for constant acceleration from rest) is the magnitude of the displacement vector.
4
Simplification of radical expressions
To simplify a square root like $\\sqrt{n}$, factor $n$ into a product of a perfect square and another integer: $n = k^2 \\times m$, where $m$ has no perfect square factors. Then $\\sqrt{n} = k\\sqrt{m}$. This is used to express the final answer in the required form $a\\sqrt{6}$.
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