Question

At time t = 0 s, an object is observed at x = 0 m; and its position along the x-axis follows this expression: x = –4t + t2, where the units for distance and time are meters and seconds, respectively. What is the object's displacement between t = 1.0 s and t = 3.0 s?

Answer 1

Answer:

The object displacement is 0 meters.

Explanation:

Because the expression for the object position as function of time is:

$x(t)= -4t + t^2$

At t= 3.0=t2 seconds the x-position of the object is:

$x(3.0)= -4(3.0) + (3.0)^2$

$x(3.0s)=-3 m$

and at t=1.0=t1 seconds the x-position of the objects is:

$x(1.0)= -4(1.0) + (1.0)^2$

$x(1.0s)=-3m$

The total displacement ($\Delta x$) of the object between t2 and t1 is the difference between x(3.0s) and x(1.0s):

$\Delta x= x(3.0)-x(1.0)=-3.0-(-3.0)$

$\Delta x=0m$

The displacement of the object is zero!!!