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Two forces, `underset~F_(text(A)) = 4 underset~i - 2 underset~j` and `underset~F_(text(B)) = 2 underset~i - 5 underset~j`, act on a particle of mass 3 kg. The particle is initially at rest at position `underset~i + underset~j`. All force components are measured in newtons and displacements are measured in metres.
The cartesian equation of the path of the particle is
`E`
`text(Net Force)\ (underset~F) = 4underset~i – 2underset~j + 2underset~i + 5underset~j = 6underset~i + 3underset~j`
`underset~a = underset~F/m = 2underset~i + underset~j`
`underset~v = int_0^t 2underset~i + underset~j\ dt = 2tunderset~i + tunderset~j`
| `underset~r` | `= int_0^t 2tunderset~i + tunderset~j\ dt + (underset~i + underset~j)` |
| `= t^2 underset~i + (t^2)/2 underset~j + underset~i + underset~j` | |
| `= (t^2 + 1)underset~i + ((t^2)/2 + 1)underset~j` |
`x = t^2 + 1 \ => \ t^2 = x – 1`
| `y` | `= (t^2)/2 + 1` |
| `= ((x – 1))/2 + 1` | |
| `= x/2 + 1/2` |
`=>E`
A particle is moving along the `x`-axis with velocity `underset~v = u underset~i`, where `u` is a real constant.
At time `t = 0`, a force acts on the particle, causing it to accelerate with acceleration `underset~a = alpha underset~j`, where `alpha` is a negative real constant.
Which one of the following statements correctly describes the motion of the particle for `t > 0`?
`E`
`underset~a = alpha underset~j`
`text(Once force acts on particle:)`
`underset~v = u underset~i + alphat underset~j,\ \ \ (text(let)\ c_1 = 0)`
`underset~r = utunderset~i + alpha/2 t^2 underset~j,\ \ \ (text(let)\ c_2 = 0)`
`underset~r\ \ text(describes a parabola.)`
`=>E`
The acceleration vector of a particle that starts from rest is given by
`underset ~a(t) = −4 sin(2t) underset ~i + 20 cos (2t) underset ~j - 20 e^(−2t) underset ~k`, where `t >= 0`.
The velocity vector of the particle, `underset ~v(t)`, is given by
`D`
| `underset ~ dot v(t)` | `= underset ~a(t)` |
| `underset ~v(t)` | `= int underset ~a (t)\ dt` |
| `= (2 cos (2t) + c_0) underset ~i + (10 sin (2t) + c_1) underset ~j + (10e^(-2t) + c_2) underset ~k` |
`text(S)text(ince)\ \ v=0\ \ text(when)\ \ t=0:`
| `0` | `= (2 cos (0) + c_0) underset ~i + (10 sin (0) + c_1) underset ~j + (10e^0 + c_2) underset ~k` |
| `0` | `= (2 + c_0) underset ~i + c_1 underset ~j + (10 + c_2) underset ~k ` |
`=> c_0 = -2, \ \ c_1 = 0, \ \ c_2 = -10`
`:. underset ~v(t) = (2 cos (2t) – 2) underset ~i + 10 sin(2t) underset ~j + (10e^(-2t) – 10) underset ~k`
`=> D`
A particle of mass 2 kg with initial velocity `3underset~i + 2underset~j` ms−1 experiences a constant force for 10 seconds.
The particle's velocity at the end of the 10-second period `43underset~i - 18underset~j` ms−1 .
| a. | `u_i` | `= 3` | `v_i` | `= 43` |
| `u_j` | `= 2` | `v_j` | `= −18` |
| `43` | `= 3 + 10a` |
| `40` | `= 10a` |
| `a_i` | `= 4` |
♦ Mean mark 38%.
| `−18` | `= 2 + 10a_j` |
| `−20` | `= 10a_j` |
| `a_j` | `= −2` |
| `underset~a` | `= 4underset~i – 2underset~j` |
| `|underset~a|` | `= sqrt(4^2 + (−2)^2)` |
| `= sqrt(20)` | |
| `= sqrt(4 xx 5)` | |
| `= 2sqrt5` |
| `:.|F|` | `= 2 xx 2sqrt5` |
| `= 4sqrt5 N` |
b. `underset~a = 4underset~i – 2underset~j`
| `underset~v` | `= int4underset~i – 2underset~j\ dt` |
| `= (4t + C_i)underset~i + (−2t + C_j)underset~j\ \ \ (C_i, C_j ∈ R)` |
`underset~v(0) = C_i underset~i + C_j underset~j = 3underset~i + 2underset~j`
`=> C_i = 3, C_j = 2`
`underset~v(t) = (4t + 3)underset~i + (2 – 2t)underset~j`
| `underset~x(t)` | `= int(4t + 3)underset~i + (2 – 2t)underset~j\ dt` |
| `= (((4t^2)/2 +3t) + b_i)underset~i + ((2t – (2t^2)/2) + b_j)underset~j \ \ \ (b_i, b_j ∈ R)` | |
| `= (2t^2 + 3t + b_i)underset~i + (2t – t^2 + b_j)underset~j` |
`underset~x(0) = b_iunderset~i + b_junderset~j`
`underset~0 => b_i = b_j = 0`
`underset~x(t) = (2t^2 + 3t)underset~i + (2t – t^2)underset~j`
| `underset~x(10)` | `= (2(10)^2 + 3(10))underset~i + (2(10) – 10^2)underset~j` |
| `= 230underset~i – 80underset~j` |
The position vector of a particle moving along a curve at time `t` is given by `underset~r(t) = cos^3(t)underset~i + sin^3(t)underset~j, \ 0 <= t <= pi/4`.
Find the length of the path that the particle travels along the curve from `t = 0` to `t = pi/4`. (4 marks)
`3/4`
`x(t) = (cos(t))^3 `♦ Mean mark 50%.
`xprime(t)= −3cos^2(t)sin(t)`
`y(t) = (sin(t))^3`
`yprime(t)= 3sin^2(t)cos(t)`
| `l` | `= int_0^(pi/4)sqrt((xprime(t))^2 + (yprime(t))^2)\ dt` |
| `= int_0^(pi/4)sqrt(9cos^4(t)sin^2(t) + 9sin^2(t)cos^2(t))\ dt` | |
| `= int_0^(pi/4)3sqrt(sin^2(t)cos^2(t)(cos^2(t) + sin^2(t)))\ dt` | |
| `= int_0^(pi/4)3sqrt(sin^2(t)cos^2(t))\ dt` | |
| `= 3int_0^(pi/4)sin(t)cos(t)\ dt` | |
| `= 3/2 int_0^(pi/4)(2sin(t)cos(t))\ dt` | |
| `= 3/2int_0^(pi/4)sin(2t)\ dt` | |
| `= 3/2[−1/2 cos(2t)]_0^(pi/4)` | |
| `= −3/4(cos(pi/2) – cos(0))` | |
| `= 3/4` |
The position vector of a particle moving along a curve at time `t` seconds is given by
`underset ~r (t) = t^3/3 underset ~i + (text{arcsin}(t) + t sqrt (1 - t^2)) underset ~j, \ 0 <= t <= 1`,
where distances are measured in metres.
The distance `d` metres that the particle travels along the curve in three-quarters of a second is given by
`d = int_0^(3/4) (at^2 + bt + c)\ dt`
Find `a, b` and `c`, where `a, b, c in Z`. (5 marks)
`a = -1,\ \ b = 0,\ \ c = 2`
| `x` | `= t^3/3` |
| `x′ (t)` | `=t^2` |
♦♦ Mean mark 26%.
| `y` | `= sin^(-1)(t) + t(1 – t^2)^(1/2)` |
| `y′(t)` | `= 1/sqrt(1 – t^2) + sqrt(1 – t^2) + t(1/2(-2t)(1 – t^2)^(-1/2))` |
| `= 1/sqrt(1 – t^2) + sqrt (1 – t^2) – t^2/sqrt(1 – t^2)` | |
| `= (1 – t^2)/sqrt (1 – t^2) + sqrt (1 – t^2)` | |
| `= sqrt(1 – t^2) + sqrt(1 – t^2)` | |
| `= 2 sqrt (1 – t^2)` |
| `d` | `= int_0^(3/4) sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2)\ dt` |
| `= int_0^(3/4) sqrt(t^4 + 4(1 – t^2))\ dt` | |
| `= int_0^(3/4) sqrt(t^4 – 4t^2 + 4)\ dt` | |
| `= int_0^(3/4) sqrt((t^2 – 2)^2)\ dt` |
`text(S)text(ince distance travelled along curve is positive,)`
`text(When)\ \ 0 <= t <= 1 => sqrt((t^2 – 2)^2) = 2 – t^2 >= 0`
`d = int_0^(3/4) – t^2 + 0*t + 2\ dt`
`a = -1,\ \ b = 0,\ \ c = 2`