Question

A Particle Moves with Constant $acc^n$

At t=0, Speed = v t=1, Speed = v/2 t-2, speed = v/4

What is speed at t = 3s ?

Answer 1

$\frac{\sqrt{7}v}{4}$

Answer 2

The problem states that a particle moves with constant acceleration and provides its speed at three different times. Let $\vec{V}(t)$ be the velocity of the particle at time $t$, and $\vec{a}$ be the constant acceleration. The initial velocity at $t=0$ is $\vec{V_0}$.

The velocity at time $t$ is given by: $\vec{V}(t) = \vec{V_0} + \vec{a}t$

The speed is the magnitude of the velocity, i.e., $|\vec{V}(t)|$. We are given:

1. At $t=0$: $|\vec{V_0}| = v$
2. At $t=1$: $|\vec{V_0} + \vec{a}| = v/2$
3. At $t=2$: $|\vec{V_0} + 2\vec{a}| = v/4$

Let's square these equations to work with dot products, which are easier for vector algebra:

1. $|\vec{V_0}|^2 = \vec{V_0} \cdot \vec{V_0} = v^2$
2. $|\vec{V_0} + \vec{a}|^2 = (\vec{V_0} + \vec{a}) \cdot (\vec{V_0} + \vec{a}) = \vec{V_0} \cdot \vec{V_0} + 2\vec{V_0} \cdot \vec{a} + \vec{a} \cdot \vec{a} = (v/2)^2 = v^2/4$
3. $|\vec{V_0} + 2\vec{a}|^2 = (\vec{V_0} + 2\vec{a}) \cdot (\vec{V_0} + 2\vec{a}) = \vec{V_0} \cdot \vec{V_0} + 4\vec{V_0} \cdot \vec{a} + 4\vec{a} \cdot \vec{a} = (v/4)^2 = v^2/16$

Substitute $\vec{V_0} \cdot \vec{V_0} = v^2$ into equations (2) and (3): From (2): $v^2 + 2\vec{V_0} \cdot \vec{a} + |\vec{a}|^2 = v^2/4$ $2\vec{V_0} \cdot \vec{a} + |\vec{a}|^2 = v^2/4 - v^2 = -3v^2/4$ (Equation A)

From (3): $v^2 + 4\vec{V_0} \cdot \vec{a} + 4|\vec{a}|^2 = v^2/16$ $4\vec{V_0} \cdot \vec{a} + 4|\vec{a}|^2 = v^2/16 - v^2 = -15v^2/16$ (Equation B)

Let $X = \vec{V_0} \cdot \vec{a}$ and $Y = |\vec{a}|^2$. We have a system of linear equations: A: $2X + Y = -3v^2/4$ B: $4X + 4Y = -15v^2/16$

Multiply Equation A by 4: $8X + 4Y = -3v^2$ (Equation C)

Subtract Equation B from Equation C: $(8X + 4Y) - (4X + 4Y) = -3v^2 - (-15v^2/16)$ $4X = -48v^2/16 + 15v^2/16$ $4X = -33v^2/16$ $X = -33v^2/64$

Now substitute $X$ back into Equation A to find $Y$: $2(-33v^2/64) + Y = -3v^2/4$ $-33v^2/32 + Y = -3v^2/4$ $Y = -3v^2/4 + 33v^2/32$ $Y = -24v^2/32 + 33v^2/32$ $Y = 9v^2/32$

So, we have $\vec{V_0} \cdot \vec{a} = -33v^2/64$ and $|\vec{a}|^2 = 9v^2/32$.

Now, we need to find the speed at $t=3$s. Let this be $S_3$. $S_3 = |\vec{V_0} + 3\vec{a}|$ $S_3^2 = |\vec{V_0} + 3\vec{a}|^2 = (\vec{V_0} + 3\vec{a}) \cdot (\vec{V_0} + 3\vec{a})$ $S_3^2 = \vec{V_0} \cdot \vec{V_0} + 6\vec{V_0} \cdot \vec{a} + 9\vec{a} \cdot \vec{a}$ $S_3^2 = v^2 + 6X + 9Y$

Substitute the values of $X$ and $Y$: $S_3^2 = v^2 + 6(-33v^2/64) + 9(9v^2/32)$ $S_3^2 = v^2 - 198v^2/64 + 81v^2/32$ $S_3^2 = v^2 - 99v^2/32 + 81v^2/32$ $S_3^2 = v^2 + (-99 + 81)v^2/32$ $S_3^2 = v^2 - 18v^2/32$ $S_3^2 = v^2 - 9v^2/16$ $S_3^2 = (16v^2 - 9v^2)/16$ $S_3^2 = 7v^2/16$

Finally, take the square root to find the speed: $S_3 = \sqrt{7v^2/16} = \frac{\sqrt{7}v}{4}$

The final answer is $\boxed{\frac{\sqrt{7}v}{4}}$.

Explanation:

1. Recognize that the problem implies motion in 2D or 3D because a 1D constant acceleration scenario leads to contradictions (specifically, $v=0$) unless $v=0$.
2. Use the vector kinematic equation $\vec{V}(t) = \vec{V_0} + \vec{a}t$.
3. Square the given speed equations to eliminate square roots and work with dot products:
   • $|\vec{V_0}|^2 = v^2$
   • $|\vec{V_0} + \vec{a}|^2 = v^2/4 \implies \vec{V_0} \cdot \vec{V_0} + 2\vec{V_0} \cdot \vec{a} + |\vec{a}|^2 = v^2/4$
   • $|\vec{V_0} + 2\vec{a}|^2 = v^2/16 \implies \vec{V_0} \cdot \vec{V_0} + 4\vec{V_0} \cdot \vec{a} + 4|\vec{a}|^2 = v^2/16$
4. Substitute $v^2$ for $\vec{V_0} \cdot \vec{V_0}$ and simplify the equations into a system of two linear equations in terms of $X = \vec{V_0} \cdot \vec{a}$ and $Y = |\vec{a}|^2$.
5. Solve the system of equations for $X$ and $Y$.
6. Calculate the square of the speed at $t=3$s using the formula $S_3^2 = |\vec{V_0} + 3\vec{a}|^2 = v^2 + 6X + 9Y$.
7. Take the square root of the result to find the speed.