SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

  • Login
  • Get Help
  • About

Calculus, 2ADV C1 EQ-Bank 8

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)

    --- 5 WORK AREA LINES (style=lined) ---

  2. On the graph below, sketch the displacement, \(x\) metres, with respect to time \(t\).   (2 marks)
     
       

    --- 0 WORK AREA LINES (style=lined) ---

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

    --- 6 WORK AREA LINES (style=lined) ---

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.    \(x=t^3-9 t^2+9 t\)

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

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

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

Filed Under: Rates of Change (Adv-2027) Tagged With: Band 3, Band 4, smc-6438-10-Motion Graphs, smc-6438-20-Polynomial Function

Copyright © 2014–2025 SmarterEd.com.au · Log in