Aussie Maths & Science Teachers: Save your time with SmarterEd
The acceleration, \(a\) ms\(^{-2}\), of a particle that starts from rest and moves in a straight line is described by \(a=1+v\), where \(v\) ms\(^{-1}\) is its velocity after \(t\) seconds.
The velocity of the particle after \( \log _e(e+1) \) seconds is
\(A\)
\(\dfrac{dv}{dt}=1+v\ \ \Rightarrow \ \dfrac{dt}{dv} = \dfrac{1}{1+v} \)
\(t= \displaystyle{\int \dfrac{1}{1+v}\ dv} = \log_e{(1+v)}+c \)
\(\text{When}\ \ t=0, v=0\ \ \Rightarrow \ c=0 \)
| \(t\) | \(=\log_e(1+v) \) | |
| \(1+v\) | \(=e^t\) | |
| \(v\) | \(=e^t-1\) |
\(\text{At}\ \ t=\log_e(e+1): \)
\(v=e^{\log_e{(e+1)}}-1 = e+1-1=e \)
\(\Rightarrow A\)
At her favourite fun park, Katherine’s first activity is to slide down a 10 m long straight slide. She starts from rest at the top and accelerates at a constant rate, until she reaches the end of the slide with a velocity of `6\ text(ms)^(-1)`.
When at the top of the slide, which is 6 m above the ground, Katherine throws a chocolate vertically upwards. The chocolate travels up and then descends past the top of the slide to land on the ground below. Assume that the chocolate is subject only to gravitational acceleration and that air resistance is negligible.
At what velocity would the chocolate need to be propelled upwards, if Katherine were to immediately slide down the slide and run to reach the chocolate just as it hits the ground?
Give your answer in `text(ms)^(-1)`, correct to one decimal place. (2 marks)
Katherine’s next activity is to ride a mini speedboat. To stop at the correct boat dock, she needs to stop the engine and allow the boat to be slowed by air and water resistance.
At time `t` seconds after the engine has been stopped, the acceleration of the boat, `a\ text(ms)^(-2)`, is related to its velocity, `v\ text(ms)^(-1)`, by
`a = -1/10 sqrt(196 - v^2)`.
Katherine stops the engine when the speedboat is travelling at `7\ text(m/s)`.
a. `u = 0, quad v = 6, quad s = 10`COMMENT: Exact form required. 3.3 seconds was marked incorrect!
| `text(Solve for)\ \ t:\ \ \ ((u + v)/2)t` | `= s` |
| `((0 + 6)/2)t` | `= 10` |
| `t` | `= 10/3\ text(s)` |
b. `u = 10, quad a = -9.8, quad s = -6`COMMENT: Many students showed a “lack of understanding” of displacement here.
`text(Solve for)\ \ t:`
| `s` | `= ut + 1/2 at^2` |
| `-6` | `= 10t – 4.9 t^2` |
| `:. t` | `~~2.5\ text(s)\ \ \ text{(by CAS)}` |
♦♦ Mean mark 27%.
c. `text(Time of chocolate in air)`
`= 10/3 + 4`
`=22/3`
`text(Solve for)\ \ u:`
| `-6` | `= 22/3 u -4.9(22/3)^2` |
| `:. u` | `~~35.1\ text(m s)^(−1)\ \ \ text{(by CAS)}` |
| d.i. | `(dv)/(dt)` | `= -1/10 sqrt(196 – v^2)` |
| `(dt)/(dv)` | `= (-10)/sqrt(196 – v^2)` | |
| `t` | `= int(-10)/sqrt(196 – v^2)\ dv` | |
| `-t/10` | `= int 1/sqrt(14^2 – v^2) dv` | |
| `-t/10` | `= sin^(-1)(v/14) + c` |
`text(When)\ \ t = 0, v = 7`
`=> c=-sin^(-1)(1/2) = -pi/6`
| `-t/10` | `=sin^(-1)(v/14) – pi/6` | |
| `sin^(-1) (v/14)` | `= pi/6-t/10` | |
| `:. v` | `=14sin(pi/6-t/10)` |
d.ii. `text(Find)\ \ t\ \ text(when)\ \ v=0:`
| `-t/10` | `=sin^(-1)(0) – pi/6` | |
| `t` | `= -10 sin^(-1)(0) + (10 pi)/6` | |
| `= (5 pi)/3\ text(s)` |
d.iii. `v = 14sin(pi/6-t/10)`
| `(dx)/(dt)` | `=14sin(pi/6-t/10)` |
| `x` | `= int_0^((5pi)/3)14sin(pi/6-t/10)\ dt` |
| `~~ 18.8\ text(m)\ \ \ text{(by CAS)}` |
A particle moves in a straight line such that its acceleration is given by `a = sqrt(v^2 - 1)` , where `v` is its velocity and `x` is its displacement from a fixed point.
Given that `v = sqrt2` when `x = 0`, the velocity `v` in terms of `x` is
A. `v = sqrt(2 + x)`
B. `v = 1 + |\ x + 1\ |`
C. `v = sqrt(2 + x^2)`
D. `v = sqrt(1 + (1 + x)^2)`
E. `v = sqrt(1 + (x - 1)^2)`
`D`
| `v(dv)/(dx)` | `= sqrt(v^2 – 1)` |
| `(dv)/(dx)` | `= sqrt(v^2 – 1)/v` |
| `(dx)/(dv)` | `= v/sqrt(v^2 – 1)` |
| `x` | `=int v/sqrt(v^2 – 1)\ dv` |
| `= sqrt(v^2 – 1) + c` |
`text(When)\ \ x=0, \ v = sqrt2\ \ =>\ \ c = −1`
| `x` | `= sqrt(v^2 – 1) -1` |
| `v^2` | `= 1 + (1 + x)^2` |
`:. v = sqrt(1 + (1 + x)^2)`
`=> D`
A ball is thrown vertically up with an initial velocity of `7sqrt6` ms`\ ^(–1)`, and is subject to gravity and air resistance.
The acceleration of the ball is given by `ddotx = −(9.8 + 0.1v^2)`, where `x` metres is its vertical displacement, and `v` ms`\ ^(–1)` is its velocity at time `t` seconds.
The time taken for the ball to reach its maximum height is
A. `pi/3`
B. `(5pi)/(21sqrt2)`
C. `log_e(4)`
D. `(10pi)/(21sqrt2)`
E. `10log_e(4)`
`D`
`(dv)/(dt) = −(9.8 + 0.1v^2)`♦ Mean mark 42%.
`(dt)/(dv) = −1/(9.8 + 0.1v^2)`
`text(Max height occurs when)\ \ v=0.`
`text(Find)\ \ t\ \ text(when)\ \ v=0:`
| `t` | `= int_(7sqrt6)^0 −1/(9.8 + v^2/10)\ dv` |
| `= (10pi)/(21sqrt2)` |
`=> D`
A helicopter is hovering at a constant height above a fixed location. A skydiver falls from rest for two seconds from the helicopter. The skydiver is subject only to gravitational acceleration and air resistance is negligible for the first two seconds. Let downward displacement be positive.
After two seconds, air resistance is significant and the acceleration of the skydiver is given by `a = g -0.01v^2`.
a. `u = 0, \ t = 2, \ a = g = 9.8`
| `x` | `= ut + 1/2 at^2` |
| `=0 xx 2 + 1/2 xx 9.8 xx 2^2` | |
| `= 19.6\ text(m)` |
| b. | `v` | `= u+ at` |
| `v(2)` | `= 9.8 xx 2` | |
| `= 19.6\ text(ms)^(−1)\ \ \ text{(downward → positive)}` |
c. `text(Terminal velocity), v_t, text(occurs when)\ \ a = 0,`♦ Mean mark 48%.
| `g -0.01v_t^2` | `=0` |
| `v_t^2` | `= 100 xx 9.8` |
| `:. v_t` | `= 14 sqrt5\ text(ms)^(−1)\ \ \ (v_t > 0\ \ text{as}\ \ v_t\ \ text{is downwards})` |
| d.i. | `(dv)/(dt)` | `= g – 0.01v^2` |
| `(dt)/(dv)` | `= 1/(g – 0.01v^2)` | |
| `= 100/(100g – v^2)` | ||
| `t` | `= int 100/(100g – v^2)` |
`text(Time taken until)\ \ v=19.6\ \ text(is 2 seconds.)`♦ Mean mark part (d)(i) 39%.
`:.\ text(Time taken until)\ \ v=30`
`=int_19.6^30 100/(100g – v^2)\ dv + 2`
d.ii. `5.8\ text(seconds)\ \ \ text{(by CAS)}`♦♦ Mean mark part (d)(ii) 25%.
| e. | `v *(dv)/(dx)` | `= g – 0.01 v^2` |
| `(dv)/(dx)` | `= g/v – (v^2)/(100v)` | |
| `= (100g – v^2)/(100v)` | ||
| `(dx)/(dv)` | `= (100v)/(100g – v^2)` | |
| `x` | `= int (100v)/(100g – v^2)\ dv` |
`text(Distance fallen in 1st 2 seconds)\ = 19.6\ text(m)`♦♦ Mean mark part (e) 29%.
`:.\ text(Total distance fallen until)\ \ v=30`
`= int_19.6^30 (100v)/(100g – v^2)\ dv + 19.6`
`~~ 120\ text(m)`
The acceleration, `a\ text(ms)^(-2)`, of a particle moving in a straight line is given by `a = v^2 + 1`, where `v` is the velocity of the particle at any time `t`. The initial velocity of the particle when at origin `O` is `2\ text(ms)^(-1)`.
The displacement of the particle from `O` when its velocity is `3\ text(ms)^(-1)` is
`C`
| `v* (dv)/(dx)` | `= v^2 + 1` |
| `(dv)/(dx)` | `= (v^2 + 1)/v` |
| `(dx)/(dv)` | `= v/(v^2 + 1)` |
| `x` | `= int (v/(v^2 + 1))\ dv` |
| `=1/2 ln(v^2 + 1)+c` |
`text(When)\ \ x=0, \ v=2:`
`c=-1/2 ln5`
`text(Find)\ \ x\ \ text(when)\ \ v=3:`
| `x` | `= 1/2ln(9 + 1) – 1/2 ln5` |
| `= 1/2 ln (10/5)` | |
| `= 1/2 ln 2` |
`=> C`
The acceleration `a` ms¯² of a body moving in a straight line in terms of the velocity `v` ms¯¹ is given by `a = 4v^2.`
Given that `v = e` when `x = 1`, where `x` is the displacement of the body in metres, find the velocity of the body when `x = 2.` (4 marks)
`e^5`
| `v* (dv)/(dx)` | `= 4v^2` |
| `(dv)/(dx)` | `= 4v` |
| `(dx)/(dv)` | `= 1/(4v)` |
| `x` | `=int 1/(4v) \ dv` |
| `=1/4 ln (4v)+c` |
`text(When)\ \ x=1, v=e`
| `1` | `=1/4 ln(4e)+c` | |
| `:.c` | `=1-1/4 ln(4e)` |
`text(Find)\ \ v\ \ text(when)\ \ x=2:`
COMMENT: Strong log and exponential calculation ability required here.
| `1/4 ln (4v)+1-1/4 ln(4e)` | `= 2` |
| `1/4 ln(4v)` | `= 1/4 ln (4e)+1` |
| `ln(4v)` | `= ln(4e)+4` |
| `e^(ln4v)` | `= e^(ln(4e) +4)` |
| `4v` | `= e^(ln(4e)) * e^4` |
| `4v` | `=4e xx e^4` |
| `:.v` | `=e^5` |