If displacement is 0, does that mean initial velocity equals final velocity?

Question

For instance, one of the kinematic equations is :

$$v_f^2 = v_i^2 + 2ad$$

where $v_f$ is final velocity, $v_i$ is initial velocity, $a$ is acceleration, and $d$ is displacement.

Say for instance a guy rides a bike in circles for hours with initial velocity of $10\ m/s$ and has an acceleration of $1\ m/s^2$, but he finished on the same spot he started on. His displacement would be $0$, right?

So the $2ad$ part of the equation would turn out to be $2 \times 1 \times 0 = 0$.

This means we're left with

$$v_f^2 = v_i^2 + 0.$$

Taking the square root of each side means the final velocity equals the initial velocity, but I stated that he sped up in the problem and thus the final velocity should be higher.

Why does this not work? Do I have to use distance instead of displacement for an equation like this?

Answer 1

As @AccidentalTaylorExpansion and @David White discuss in their answers, your relationship is only valid for constant linear acceleration. Your situation is rotational, not linear motion. Also, acceleration is a vector and is not constant for circular motion.

For circular rotational motion of a particle about a fixed axis with constant magnitude angular acceleration, $\alpha$, the appropriate relationship is $\omega^2_f = \omega^2_i + 2 \alpha \theta$ where $\omega$ is angular speed equal to $d \theta/dt$ and $\theta$ is angular displacement. $\alpha = d \omega/dt$ and unless $\alpha$ is zero $\omega $ is not constant. For circular motion, the particle returns to the same position in space once per revolution, even though $\theta$ is constantly increasing.

Answer 2

The kinematic equations only apply to situations where the acceleration is constant. In your example the guy on is bike is accelerating at different rates so the kinematic equations do not apply. That's why they refer to $a$ as the acceleration.

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Edit: after the change of the post the situation now refers to constant acceleration but along a closed path (you end up where you started).

In that case the formula would be correct. In this case $d$ measures the traversed distance so at your final position $d$ would be equal to the total path you travelled.

This situation is a bit more complicated and you have to be careful here. Now $a$ is acceleration in the direction of your velocity. Because you're travelling in a circle you will also have acceleration in the sideways direction but you have to ignore this for this problem. Why is that? This is because of the work energy theorem: $$\tfrac 1 2 m v_f^2=\tfrac 1 2 mv_i^2+\int dxF_{par}(x)$$ which you can also write as $$ v_f^2=v_i^2+\frac{2}{m}\int dxF_{par}(x)$$ where $F_{par}$ is the force parallel to the velocity. When the path is a straight line and the acceleration is constant you get $\frac{2}{m}\int dx F_{par}=\frac{2}{m}\int dx\, ma=\frac{2}{m}[max]_{x=0}^{x=d}=2ad$. This gives the kinematic equation. In case this last bit was confusing I invite you to totally ignore it because it probably uses some concepts you never heard of. I included it in case you were wondering where it came from.

Answer 3

The kinematic equations were derived with straight-line motion and constant acceleration in mind. When your hypothetical bicycle rider is riding in a circle, he is never traveling in a straight line. In addition, acceleration is a vector, and in the case of circular motion, acceleration is constantly changing because its direction is constantly changing, so acceleration is never constant when traveling in circular motion. This means that the kinematic equations are invalid for circular motion, and the equations of circular motion should be used instead.

Answer 4

Taking the square root of each side means the final velocity equals the initial velocity

Not quite -- the final speed (magnitude of velocity) equals the initial speed. It could be $v_f = -v_i$.

Your original equation is for 1D motion with constant acceleration (constant in magnitude and direction), so it doesn't directly apply to motion in a circle.

However, there is a broader case where "returning to the original position means returning to the original speed": any conservative force. For example, a mass on a spring without friction will likewise have the same speed after a net displacement of zero, even though the acceleration is not constant. This generalization applies due to conservation of energy. A constant force (like gravity near the Earth's surface) is a special case of a conservative force.