kinematics

• Sep 9th 2008, 07:07 AM
raymond lim
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
• Sep 9th 2008, 08:39 AM
Simplicity
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
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
(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
(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.
• Sep 9th 2008, 07:40 PM
raymond lim
Thank you AIR