Question 1
The position $x$ (m) at time $t$ (s) of a $7$ kg particle moving in a straight line is given by
$$x=3t^3-5t^2+2t-4\ \text{ for }\ 0\le t\le10$$
Determine the time when the particle has a momentum of $620\ \text{kg m s}^{-1}$.
Not quite. $1.73$ solves $3t^3-5t^2+2t-4=0$, which is when the particle passes through the origin. Its velocity then is only about $11.6\ \text{m s}^{-1}$, so its momentum is about $81\ \text{kg m s}^{-1}$. Momentum comes from the velocity: solve $7\frac{dx}{dt}=620$.
Not quite. $2.60$ is the size of the negative solution. The equation $9t^2-10t+2=\frac{620}{7}$ has two solutions, $t\approx-2.595$ and $t\approx3.706$. The domain is $0\le t\le10$, so the negative solution is rejected altogether. Dropping its minus sign doesn’t turn it into a time the particle reaches.
Not quite. At $t=3.66$ the velocity is $9(3.66)^2-10(3.66)+2\approx85.96\ \text{m s}^{-1}$, so the momentum is about $7\times85.96\approx602\ \text{kg m s}^{-1}$, short of $620$. $3.66$ is the time for a momentum of $602$, so check you have copied $620$ the right way round. Solving $7(9t^2-10t+2)=620$ gives $t\approx3.706$.
That’s it. Step through the hints below to see why.
Momentum $p$ in $\text{kg m s}^{-1}$ against time in seconds. The curve is $p=7(9t^2-10t+2)$, drawn for $-3\le t\le5$ so you can see both places it meets the blue line. The grey band is $t<0$, outside the domain $0\le t\le10$.
At $t=3.71$ the momentum is about $621\ \text{kg m s}^{-1}$, as close to $620$ as a time to two decimal places gets. The exact solution is $t\approx3.706$. This is option (D).
At $t=3.66$ the momentum is only about $602\ \text{kg m s}^{-1}$. It hasn’t reached $620$ yet, so (C) is a little too early.
At $t=1.73$ the particle is passing through the origin, so $x\approx0$. Its momentum is only about $81\ \text{kg m s}^{-1}$. Solving $x=0$ answers a different question.
At $t=2.60$ the momentum is only about $258\ \text{kg m s}^{-1}$. The number $2.60$ is the size of the other solution, $t\approx-2.60$, the hollow dot. That time is outside the domain.
The orange point is below the blue line, so the momentum is less than $620$ here. Move later in time.
The orange point is above the blue line, so the momentum is more than $620$ here. Move earlier in time.
Momentum is mass times velocity, $p=mv$. The mass is $7$ kg, so you need the time when the velocity is $\frac{620}{7}$.
$$7v=620\ \Rightarrow\ v=\frac{620}{7}\approx88.57\ \text{m s}^{-1}$$
You are given the position, so differentiate it to get the velocity.
$$v=\frac{dx}{dt}=9t^2-10t+2$$
Solve the equation with your GDC’s solver, or as a quadratic.
$$9t^2-10t+2=\frac{620}{7}$$
$$t\approx-2.595\ \text{ or }\ t\approx3.706$$
Only the positive solution is in the domain $0\le t\le10$.
Momentum is $p=mv$, and the velocity comes from differentiating the position.
$$v=\frac{dx}{dt}=9t^2-10t+2$$
$$7\left(9t^2-10t+2\right)=620$$
$$t\approx3.706\quad(t\ge0)$$
The particle has a momentum of $620\ \text{kg m s}^{-1}$ when $t\approx3.71$ s.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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