Particle has an initial positive velocity & travels 30m in the 4th second.
This is my working out so far, but I'm not sure how to go from there. Any help on how to go about this problem would be appreciated , thanks in advance!
Given a = 3t+2:
When t=4, x (displacement)=30
v= ∫ a dt
= ∫ 3t +2 dt
= 1.5t^2 + 2t + c
x (displacement) = ∫ velocity dt
= 0.5t^3 + t^2 + ct + g (where g is another constant)
when t=4:
How do you deduce the constants c and g?
This question is from Sadler Methods Unit 3 (Year 12).
Best Answer
$a=3t+2 \Rightarrow v = \dfrac{3}{2}t^2+2t+v_0 \Rightarrow x=\dfrac{1}{2}t^3+t^2+v_0t+x_0$
$v_0$ and $x_0$ are respectively the initial velocity and position.
You want $v(5) = \dfrac{75}{2}+10+v_0=\dfrac{95}{2}+v_0$.
You know $x(4)-x_0=30 \, \mathrm{m}$ because you said that the body has travelled $30 \, \mathrm{m}$ in 4 seconds (from the inital time $t=0 \ \mathrm{s}$).
So $64/2+16+4v_0+x_0-x_0=30\Rightarrow 18 = 4v_0 \Rightarrow v_0 = \dfrac{18}{4} = \dfrac{9}{2}$.
Therefore $v(5) = \dfrac{95}{2} + \dfrac{9}{2} = \dfrac{104}{2} = 52 \, \mathrm{m.s^{-1}}$.
I hope it helps you.