Given the kinematics equation for the motion of an object falling from rest, x=.

Question

Given the kinematics equation for the motion of an object falling from rest, x=.5g*t2), what kind of relationship is predicted between x and t? (select all that apply) Question options: x=k*t+b, where k and b are constants. displacement, x, has a linear relationship with time, t. displacement, x, and time, t, obey a power law. x=k*t2, where k is a constant. displacement, x, is proportional to time, t. diplacement, x, is proportional to the square of the time, t2. x=k*t, where k is a constant Given the kinematics equation for the motion of an object falling from rest, x=.5g*t2), what kind of relationship is predicted between x and t? (select all that apply) Given the kinematics equation for the motion of an object falling from rest, x=.5g*t2), what kind of relationship is predicted between x and t? (select all that apply) x=k*t+b, where k and b are constants. displacement, x, has a linear relationship with time, t. displacement, x, and time, t, obey a power law. x=k*t2, where k is a constant. displacement, x, is proportional to time, t. diplacement, x, is proportional to the square of the time, t2. x=k*t, where k is a constant x=k*t+b, where k and b are constants. displacement, x, has a linear relationship with time, t. displacement, x, and time, t, obey a power law. x=k*t2, where k is a constant. displacement, x, is proportional to time, t. diplacement, x, is proportional to the square of the time, t2. x=k*t, where k is a constant x=k*t+b, where k and b are constants. displacement, x, has a linear relationship with time, t. displacement, x, and time, t, obey a power law. x=k*t2, where k is a constant. displacement, x, is proportional to time, t. diplacement, x, is proportional to the square of the time, t2. x=k*t, where k is a constant

Explanation / Answer

From the kinematics relation S = ut + ( 1/ 2) at 2 Where S = displacment             u = Initial velocity               = 0     Since it fall from rest             t = time            a = accleration               = + g Substitute values we get S = 0 + ( 1/ 2) gt 2                                       S = ( 1/ 2) gt 2                                          = k t 2    Where k = constant                = g / 2                                          = k t 2    Where k = constant                = g / 2 i.e., diplacement, x, is proportional to the square of the time, t2

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