Question

Which of the following equations denote motion with constant acceleration?

$\square$ v = 3t

$\square$ x = 4t$^2$ + 3t

$\square$ v = 4$\sqrt{x}$

$\square$ t = $\sqrt{x}$

• A. v = 3t
• B. x = 4t$^2$ + 3t
• C. v = 4$\sqrt{x}$
• D. t = $\sqrt{x}$

Answer 1

Answer: (A), (B), (C), (D)

All of the above

Answer 2

To determine which equations denote motion with constant acceleration, we need to find the acceleration ($a$) for each equation and check if it is a constant value (independent of time, position, or velocity).

Recall the definitions:

• Velocity: $v = \frac{dx}{dt}$
• Acceleration: $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$
• Alternatively, $a = v \frac{dv}{dx}$

Let's analyze each option:

1. $v = 3t$ This equation gives velocity as a function of time. To find acceleration, differentiate $v$ with respect to $t$: $a = \frac{dv}{dt} = \frac{d}{dt}(3t) = 3$ Since $a = 3$ (a constant), this equation represents motion with constant acceleration.

2. $x = 4t^2 + 3t$ This equation gives position as a function of time. First, find velocity by differentiating $x$ with respect to $t$: $v = \frac{dx}{dt} = \frac{d}{dt}(4t^2 + 3t) = 8t + 3$ Now, find acceleration by differentiating $v$ with respect to $t$: $a = \frac{dv}{dt} = \frac{d}{dt}(8t + 3) = 8$ Since $a = 8$ (a constant), this equation represents motion with constant acceleration.

3. $v = 4\sqrt{x}$ This equation gives velocity as a function of position. We can use the formula $a = v \frac{dv}{dx}$. First, find $\frac{dv}{dx}$: $v = 4x^{1/2}$ $\frac{dv}{dx} = \frac{d}{dx}(4x^{1/2}) = 4 \cdot \frac{1}{2} x^{(1/2)-1} = 2x^{-1/2} = \frac{2}{\sqrt{x}}$ Now, calculate acceleration: $a = v \frac{dv}{dx} = (4\sqrt{x}) \cdot \left(\frac{2}{\sqrt{x}}\right) = 8$ Since $a = 8$ (a constant), this equation represents motion with constant acceleration. Alternatively, squaring both sides of $v = 4\sqrt{x}$ gives $v^2 = 16x$. This is of the form $v^2 = u^2 + 2a(x-x_0)$, which for $u=0, x_0=0$ becomes $v^2 = 2ax$. Comparing $v^2 = 16x$ with $v^2 = 2ax$, we get $2a = 16$, so $a = 8$.

4. $t = \sqrt{x}$ This equation gives time as a function of position. It's easier to first express position ($x$) as a function of time ($t$). From $t = \sqrt{x}$, we square both sides to get $x = t^2$. Now, find velocity by differentiating $x$ with respect to $t$: $v = \frac{dx}{dt} = \frac{d}{dt}(t^2) = 2t$ Now, find acceleration by differentiating $v$ with respect to $t$: $a = \frac{dv}{dt} = \frac{d}{dt}(2t) = 2$ Since $a = 2$ (a constant), this equation represents motion with constant acceleration.

All four given equations represent motion with constant acceleration.