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Calculus, 2ADV C1 EQ-Bank 14

A drone travels vertically from its launch pad.

It's height above ground, \(h\) metres, at time \(t\) minutes is modelled by

\(h(t)=-0.2 t^3+3 t^2+5 t\)  for  \(0 \leq t \leq 12\)

  1. Find the velocity of the drone at time \(t\) minutes.   (1 mark)

  2. Determine the exact time interval during which the drone is descending.   (2 marks)

Show Answers Only

a.    \(\dfrac{dh}{dt}=-0.6 t^2+6 t+5\)

b.    \(\dfrac{15+10 \sqrt{3}}{3}<t \leqslant 12\)

Show Worked Solution

a.    \(h=-0.2 t^3+3 t^2+5 t\)

\(\text{Velocity of the drone}=\dfrac{d h}{d t}.\)

\(\dfrac{dh}{dt}=-0.6 t^2+6 t+5\)
 

b.    \(\text{Drone is descending when} \ \ \dfrac{dh}{dt}<0:\)

\(-0.6 t^2+6 t+5\) \(<0\)  
\(0.6 t^2-6 t-5\) \(>0\)  
\(6 t^2-60 t-50\) \(>0\)  

 
\(\text{Solve}\ \ 6 t^2-60 t-50=0:\)

\(t=\dfrac{60 \pm \sqrt{(-60)^2+4 \times 6 \times 50}}{2 \times 6}=\dfrac{60 \pm \sqrt{4800}}{12}=\dfrac{15 \pm 10 \sqrt{3}}{3}\)

 
\(\text{Since parabola is concave up:}\)

\(6 t^2-60 t-50>0\ \ \text{when}\ \ t>\dfrac{15+10 \sqrt{3}}{3} \quad\left( t=\dfrac{15-10 \sqrt{3}}{3}<0\right)\)

\(\therefore \text{Drone is descending for} \ \ \dfrac{15+10 \sqrt{3}}{3}<t \leqslant 12\)

Calculus, 2ADV C1 EQ-Bank 15

The displacement \(x\) metres from the origin at time, \(t\) seconds, of a particle travelling in a straight line is given by

\(x=t^3-9 t^2+9 t, \quad t \geqslant 0\)

  1. Find the time(s) when the particle is at the origin.   (2 marks)
  2. On the graph below, sketch the displacement, \(x\) metres, with respect to time \(t\).   (2 marks)
     
       

  3. Find the velocity of the particle when  \(t=2\).   (2 marks)

Show Answers Only

a.  \(\text{Particle at origin when}\ \ t=0, t=3.\)

b.
       
 

c.   \(\dot{x}=-15\  \text{m s}^{-1}\)

Show Worked Solution

a.     \(x\) \(=t^3-9 t^2+9 t\)
    \(=t\left(t^2-9 t+9\right)\)
    \(=t(t-3)^2\)

 
\(\text{Particle at origin when}\ \ t=0, t=3.\)

 
b.
       
 

c.    \(x=t^3-9 t^2+9 t\)

\(\dot{x}= \dfrac{dx}{dt} = 3 t^2-18 t+9\)

\(\text {When } t=2:\)

\(\dot{x}=3 \times 2^2-18 \times 2+9=-15\  \text{m s}^{-1}\)

Calculus, 2ADV C1 EQ-Bank 17

The displacement `x` metres from the origin at time `t` seconds of a particle travelling in a straight line is given by

`x = 2t^3-t^2-3t + 11`  when  `t >= 0`

  1.  Calculate the velocity when  `t = 2`.   (1 mark)

  2.  When is the particle stationary?   (2 marks)

Show Answers Only

 a.    `17\ text(ms)^(−1)`

 b.    `(1 + sqrt19)/6`

Show Worked Solution

a.   `x =2t^3-t^2-3t + 11` 

`v = (dx)/(dt) = 6t^2-2t-3`

 
`text(When)\ t = 2:`

`v= 6 xx 2^2-2 · 2-3= 17\ text(ms)^(−1)`
 

b.   `text(Particle is stationary when)\ \ v = 0`

`6t^2-2t-3=0`

`:. t` `= (2 ±sqrt((−2)^2-4 · 6 · (−3)))/12`
  `= (2 ± sqrt76)/12`
  `= (1 ± sqrt19)/6`
  `= (1 + sqrt19)/6 qquad(t >= 0)`