1. A particle is moving in a straight line, starting from the origin . At time t seconds, the particle has a displacement of x m from the origin and a velocity v m/s. The displacement is given by x=2t-3ln(t+1). Find the distance travelled by the particle in the first 3 seconds, correct to 2 dp.
2. A certain particle moves along the x-axis in accordance with the law t=2x^2-5x+3, where x is measured in cm and t in seconds. Initially the particle is 1.5cm to the right of the origin O and moving away from O. Find an expression for the acceleration a cm/s/s in terms of x. Then describe the motion of the particle as I'm unsure what the question is looking for here. My guess is: moving away from \(O\) is the positive direction of motion with a positive decreasing velocity and acceleration (that is, \(v\to0\) and \(a\to0\) as \(t\to\infty\)).
Question 1I'll present you with two methods.
Method 1First, we need to determine if the particle changes direction any time in the first 3 seconds. We can do this by solving \(\dfrac{dx}{dt}=0\).\[\frac{dx}{dt}=2-\frac{3}{t+1}\overset{\text{set}}{=}0\implies t=\frac{1}{2}.\] So, in the first 0.5 seconds, the particle travels from position \(x(0)=0\ \text{m}\) to position \(x(1/2)=1-3\ln(3/2)\ \text{m}\), covering a distance of \(d_1=3\ln(3/2)-1\ \text{m}\).
Then in the next 2.5 seconds, the particle travels from position \(x(1/2)=1-3\ln(3/2)\ \text{m}\) to position \(x(3)=6-3\ln(4)\ \text{m}\), covering a distance of \(d_2=6-3\ln(4)-(1-3\ln(3/2))=5-3\ln(8/3)\ \text{m}\).
Thus, the total distance traveled is \(D=d_1+d_2=4-6\ln(4/3)\approx2.27\ \text{m}\ \ (2\text{DP})\).
Method 2Alternatively, we could find \[D=\int_0^3\left|2-\frac{3}{t+1}\right|dt.\] Since \(\dfrac{dx}{dt}<0\) for \(0<t<\dfrac12\), and \(\dfrac{dx}{dt}>0\) for \(t>\dfrac12\), \begin{align*}D&=\int_0^{1/2}\left(\frac{3}{t+1}-2\right)dt+\int_{1/2}^3\left(2-\frac{3}{t+1}\right)dt\\
&=x(0)-x\left(\frac12\right)+x(3)-x\left(\frac12\right)\\
&\approx 2.27\ \text{m}.\end{align*}
Question 2I would use implicit differentiation here. \[1=4x\frac{dx}{dt}-5\frac{dx}{dt}\implies \frac{dx}{dt}=\frac{1}{4x-5}.\] Then, using implicit differentiation again, \[a=\frac{d^2x}{dt^2}=\frac{-4}{(4x-5)^2}\frac{dx}{dt}=\frac{-4}{(4x-5)^3}.\]
I'll let you describe the motion of the particle.
Edit: just been informed that implicit differentiation is in the MX2 course. (I should actually read the study designs before helping in the HSC boards lol). Here's another method to Question 2.\[t=2x^2-5x+3\implies \frac{dt}{dx}=4x-5\implies v=\frac{dx}{dt}=\frac{1}{4x-5}.\] Now, using the fact that \(a=\dfrac{d}{dx}\left(\dfrac12 v^2\right)\),
\[a=\frac{d}{dx}\left[\frac{1}{2(4x-5)^2}\right]=\frac{-4}{(4x-5)^3.}\]
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