Results 1 to 3 of 3

Math Help - kinematics

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    2

    kinematics

    A particle is projected vertically upwards from the ground so that its velocity (in m/s) is given by v1=128-32t (t lies between and including 0 to 4), and v2=3t^2-32t+80 where t>4, where t is the time in seconds after projection. Calculate
    (i) the height of the particle when t=4,
    (ii) the height of the particle when t=6,
    (iii) the value of t when the acceleration of the particle is 40 m per sec per sec.

    I can solve only part (i).

    Answers given: (i) 256 m, (ii) 248 m and (iii) 12
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    You said that you have done question (i) so I assume that you won't need to be give all steps so I will explain how to conduct the question.

    Quote Originally Posted by raymond lim View Post
    A particle is projected vertically upwards from the ground so that its velocity (in m/s) is given by v1=128-32t (t lies between and including 0 to 4), and v2=3t^2-32t+80 where t>4, where t is the time in seconds after projection. Calculate
    (i) the height of the particle when t=4,
    Displacement is the integral of velocity. For t=4, you will need to integrate v_1 = 128 - 32t to get s_1=.... You know that when t=0, s=0 so you will able to find the value of the constant. After the equation of s_1 is found, you can substitute t=4 to find the displacement at that time.

    Quote Originally Posted by raymond lim View Post
    (ii) the height of the particle when t=6,
    Same principle but this time we integrate v_2 = 3t^2 - 32t + 80 as we want to find t=6 and this equation is valid for t>4 seconds. You will get a constant so to find it's value, we use the conditions. We know that when t=4, s= 256. After the equation of s_2 is found, you can substitute t=6 to find the displacement at that time.

    Quote Originally Posted by raymond lim View Post
    (iii) the value of t when the acceleration of the particle is 40 m per sec per sec.

    I can solve only part (i).

    Answers given: (i) 256 m, (ii) 248 m and (iii) 12
    Acceleration is the derivative of velocity. The derivative of v_1 give a constant value of -32 \ \mathrm{ms}^{-2} implying that for 4 seconds it travelled at the same acceleration so we do not consider that equation. Calculate the derivative of v_2 and equate this to 40 \ \mathrm{ms}^{-2}, this will lead to one unknown value ( t) which is the variable that you are looking for.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2008
    Posts
    2
    Thank you AIR
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. kinematics
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: January 20th 2010, 11:18 PM
  2. Kinematics
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: May 30th 2008, 11:39 AM
  3. Kinematics
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: April 15th 2008, 11:14 AM
  4. Kinematics... please help
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 25th 2007, 07:01 AM
  5. Need help with Kinematics !
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: August 30th 2005, 06:02 AM

Search Tags


/mathhelpforum @mathhelpforum