1. Given motion of the two particles

• Particle 1 moves along the $x$-axis: $x(t)=10+8t-3t^2$
• Particle 2 moves along the $y$-axis: $y(t)=5-8t^3$

We need the speed of particle 2 as measured in the frame of particle 1 at $t=1\,\text{s}$.

That means we need the relative velocity: $\vec v_{21}=\vec v_2-\vec v_1$

2. Velocity of particle 1

Since particle 1 moves only along the $x$-axis, $v_1=\frac{dx}{dt}=\frac{d}{dt}(10+8t-3t^2)=8-6t$

At $t=1$, $v_1=8-6(1)=2\,\text{m/s}$

So, $\vec v_1=2\hat i$

3. Velocity of particle 2

Since particle 2 moves only along the $y$-axis, $v_2=\frac{dy}{dt}=\frac{d}{dt}(5-8t^3)=-24t^2$

At $t=1$, $v_2=-24\,\text{m/s}$

So, $\vec v_2=-24\hat j$

4. Relative velocity of particle 2 with respect to particle 1

$\vec v_{21}=\vec v_2-\vec v_1=(-24\hat j)-(2\hat i)=-2\hat i-24\hat j$

Its speed is the magnitude: $|\vec v_{21}|=\sqrt{(-2)^2+(-24)^2}$ $=\sqrt{4+576}$ $=\sqrt{580}$

Given that the speed is $\sqrt v$, we get $\sqrt v=\sqrt{580}$

Therefore, $v=580$

5. Comparison with stored answer

Stored correct answer: $580$

Our derived answer is also $580$, so it agrees.