Question

try it! (continued)

1. solve for the unknown

use the relationship among displacement, velocity, and time interval to find δx₁ during δt₁ = 10.0 s.
use the same relationship to find δx₂ during δt₂ = 20.0 s.

1. evaluate the answer

. do the signs make sense?
. is the magnitude realistic?

Explanation:

Step1: Recall the formula

The relationship between displacement $\Delta x$, velocity $v$ and time - interval $\Delta t$ is $\Delta x=v\Delta t$. But since the velocity $v$ is not given, we'll assume a general - case solution for now.

Step2: Calculate $\Delta x_1$

For $\Delta t_1 = 10.0\ s$, $\Delta x_1=v\times\Delta t_1 = 10v$.

Step3: Calculate $\Delta x_2$

For $\Delta t_2 = 20.0\ s$, $\Delta x_2=v\times\Delta t_2 = 20v$.

Step4: Evaluate the signs

The sign of $\Delta x$ depends on the sign of $v$. If $v>0$, $\Delta x_1$ and $\Delta x_2$ are positive, meaning the object is moving in the positive direction. If $v < 0$, $\Delta x_1$ and $\Delta x_2$ are negative, meaning the object is moving in the negative direction.

Step5: Evaluate the magnitude

The magnitude of $\Delta x$ is proportional to the magnitude of $v$ and the time - interval. If $v$ is a reasonable velocity (for example, in the range of common speeds of objects like cars, people walking etc.), then the magnitudes of $\Delta x_1$ and $\Delta x_2$ will be realistic.

Answer:

$\Delta x_1 = 10v$, $\Delta x_2 = 20v$