Question

A student slides a book across a desk, with a velocity of +8 m/s. When her friend catches the book, it has a velocity of +7.4 m/s. If the acceleration was a constant -5.6 m/s2, how wide is the desk?

A. 0.05 m
B. 0.60 m
C. 0.83 m
D. 1.65 m

Answer 1

Because acceleration was constant throughout the slide, we can show the slide lasted

$a=\dfrac{\Delta v}{\Delta t}\iff-5.6\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{7.4\,\frac{\mathrm m}{\mathrm s}-8\,\frac{\mathrm m}{\mathrm s}}{\Delta t}$

$\implies\Delta t=0.107\,\mathrm s$

Also because accleration was constant, we know the average velocity of the book was

$\bar v=\dfrac{7.4\,\frac{\mathrm m}{\mathrm s}+8\,\frac{\mathrm m}{\mathrm s}}2=7.7\,\dfrac{\mathrm m}{\mathrm s}$

Average velocity is also given by

$\bar v=\dfrac{\Delta x}{\Delta t}\iff7.7\,\dfrac{\mathrm m}{\mathrm s}=\dfrac{\Delta x}{0.107\,\mathrm s}$

so the width of the desk must have been

$\Delta x=0.824\,\mathrm m$

which means C is the most likely answer.