Question

a skateboarder starts from rest and moves down a hill with constant acceleration in a straight line for 6 s. in a second trial he starts from rest and moves along the same straight line with the same acceleration for only 2 s. how does his displacement from his starting point compare with the first trial?
A) one-third as large
B) three times larger
C)One-ninth as large
D) nine times larger

Answer 1

Let's look at this 1D kinematics equation:\[x=x_0+v_0t+\frac{1}{2}at^2\]The displacement is defined as the final x position minus the initial x position, so we can rewrite that equation as:\[x-x_0=v_0t+\frac{1}{2}at^2 = d\]We know that the initial velocity is 0 (at rest), so we can cross that term off and write the relationship like this:\[d=\frac{1}{2}at^2\]Since both a and 1/2 are not functions of either d or t, we can drop them off when making a relationship between the two:\[d\ \alpha\ t^2\]Let's use this relationship directly in the two examples given:

1)\[d=t^2 \rightarrow d=(6)^2 \rightarrow d=36\]2)\[d=t^2 \rightarrow d=(2)^2 \rightarrow d=4\]36 is nine times that of 4:\[\frac{36}{4}=9\]This means that the displacement in the second trial will be one-ninth as large as the displacement in the first trial.