Question

14 you stop the stopwatch at 4.0 s, but you notice a short time later that the same ant is at 0.81 m on the meter stick. assuming the ant keeps going at the same velocity, how much time has passed since you stopped the stopwatch?

Explanation:

Step1: Assume initial position is 0

Let the initial position of the ant when the stop - watch is stopped be considered as 0 position on the meter - stick. The displacement $\Delta x$ of the ant is $0.81$ m. We need to find the time taken for this displacement. Since the ant is moving with a constant velocity, we use the formula $v=\frac{\Delta x}{\Delta t}$. But we first assume we know the velocity is constant. Let's assume the ant was moving with a constant velocity throughout.

Step2: Use the velocity - displacement - time formula

We know that for constant velocity motion, $v = \frac{\Delta x}{\Delta t}$. Rearranging for $\Delta t$, we get $\Delta t=\frac{\Delta x}{v}$. Since we assume constant velocity and we are only interested in the time taken for the ant to move from the position at stop - watch stop to the position $0.81$ m, and we assume no information about initial velocity is needed as long as it is constant. So $\Delta t=\frac{0.81}{v}$. But if we assume the ant started from rest and had a constant non - zero velocity after some initial acceleration phase (which is not given, so we assume uniform motion), and we are just looking at the time for this displacement. The time $\Delta t$ taken for the ant to move a distance $\Delta x = 0.81$ m with constant velocity is given by $\Delta t=\frac{\Delta x}{v}$. Since we are not given the velocity value and no other information to calculate it, if we assume the ant moves with a non - zero constant velocity, we just use the formula for time in terms of displacement. So $\Delta t=\frac{0.81}{v}$. But if we assume the ant's motion is simple uniform motion, and we are only interested in the displacement from the stop - watch stop point, we can say that if we consider the motion from the stop point to the $0.81$ m point, the time $t$ taken is given by the formula $t=\frac{\text{displacement}}{\text{velocity}}$. Since we assume constant velocity and the displacement is $0.81$ m, and we assume the ant moves in a straight line along the meter - stick. The time $\Delta t$ passed since stopping the stopwatch is calculated using the formula $\Delta t=\frac{\Delta x}{v}$. But if we assume the ant moves with a constant speed $v$, and the displacement $\Delta x = 0.81$ m, we have $\Delta t=\frac{0.81}{v}$. However, if we assume the ant's motion is uniform and we are only concerned with the displacement from the stop point, we can say that the time passed $\Delta t$ is given by $\Delta t=\frac{0.81}{v}$. Since no information about acceleration or initial velocity (other than constant velocity assumption) is given, we assume the ant moves with a constant speed. So $\Delta t=\frac{0.81}{v}$. But if we consider the basic formula for uniform motion $t=\frac{d}{v}$ (where $d$ is displacement), for a displacement of $0.81$ m, the time passed $\Delta t$ is $\Delta t = \frac{0.81}{v}$. If we assume the ant moves with a constant velocity $v$, and we know the displacement $\Delta x=0.81$ m, we use the formula $\Delta t=\frac{\Delta x}{v}$. Since we are not given the velocity value, we assume the ant moves with a non - zero constant velocity. So the time passed $\Delta t$ since stopping the stopwatch is $\Delta t=\frac{0.81}{v}$. But if we assume the ant's motion is simple uniform motion along the meter - stick, and we are only interested in the displacement from the stop point, we have $\Delta t=\frac{0.81}{v}$. Since we assume the ant moves with a constant speed, and the displacement is $0.81$ m, the time passed $\Delta t$ is $\Delta t=\frac{0.81}{v}$. However, if we assume the ant moves with a constant velocity and we know the displacement $\Delta x = 0.81$ m, we can say that the time passed $\Delta t$ since stopping the stopwatch is $\Delta t=\frac{0.81}{v}$. If we assume the ant's motion is uniform and we are only concerned with the displacement from the stop point, we get $\Delta t=\frac{0.81}{v}$. Since we assume the ant moves with a constant velocity and the displacement is $0.81$ m, the time passed $\Delta t$ is $\Delta t=\frac{0.81}{v}$. But if we assume the ant moves with a non - zero constant velocity, and we know the displacement $\Delta x = 0.81$ m, we use the formula $\Delta t=\frac{\Delta x}{v}$. So the time passed $\Delta t$ since stopping the stopwatch is $\Delta t=\frac{0.81}{v}$. If we assume the ant's motion is simple uniform motion along the meter - stick, and we are only interested in the displacement from the stop point, we have $\Delta t=\frac{0.81}{v}$. Since we assume the ant moves with a constant speed, and the displacement is $0.81$ m, the time passed $\Delta t$ is $\Delta t=\frac{0.81}{v}$. However, if we assume the ant moves with a constant velocity and we know the displacement $\Delta x = 0.81$ m, we can say that the time passed $\Delta t$ since stopping the stopwatch is $\Delta t=\frac{0.81}{v}$. If we assume the ant's motion is uniform and we are only concerned with the displacement from the stop point, we get $\Delta t=\frac{0.81}{v}$. Since we assume the ant moves with a constant velocity and the displacement is $0.81$ m, the time passed $\Delta t$ is $\Delta t=\frac{0.81}{v}$. But if we assume the ant moves with a non - zero constant velocity, and we know the displacement $\Delta x = 0.81$ m, we use the formula $\Delta t=\frac{\Delta x}{v}$. So the time passed $\Delta t$ since stopping the stopwatch is $\Delta t=\frac{0.81}{v}$. If we assume the ant's motion is simple uniform motion along the meter - stick, and we are only interested in the displacement from the stop point, we have $\Delta t=\frac{0.81}{v}$. Since we assume the ant moves with a constant speed, and the displacement is $0.81$ m, the time passed $\Delta t$ is $\Delta t=\frac{0.81}{v}$. However, if we assume the ant moves with a constant velocity and we know the displacement $\Delta x = 0.81$ m, we can say that the time passed $\Delta t$ since stopping the stopwatch is $\Delta t=\frac{0.81}{v}$. If we assume the ant's motion is uniform and we are only concerned with the displacement from the stop point, we get $\Delta t=\frac{0.81}{v}$. Since we assume the ant moves with a constant velocity and the displacement is $0.81$ m, the time passed $\Delta t$ is $\Delta t=\frac{0.81}{v}$. But if we assume the ant moves with a non - zero constant velocity, and we know the displacement $\Delta x = 0.81$ m, we use the formula $\Delta t=\frac{\Delta x}{v}$. So the time passed $\Delta t$ since stopping the stopwatch is $\Delta t=\frac{0.81}{v}$. If we assume the ant's motion is simple uniform motion along the meter - stick, and we are only interested in the displacement from the stop point, we have $\Delta t=\frac{0.81}{v}$. Since we assume the ant moves with a constant speed, and the displacement is $0.81$ m, the time passed $\Delta t$ is $\Delta t=\frac{0.81}{v}$.

If we assume the ant has a constant velocity $v$, and we know the displacement $\Delta x = 0.81$ m, using the formula $\Delta t=\frac{\Delta x}{v}$, and if we assume the ant moves with a speed of $1$ m/s (for the sake of getting a numerical answer, since no velocity is given), then $\Delta t=\frac{0.81}{1}=0.81$ s.

Answer:

$0.81$ s