Question

Haley is driving down a straight highway at 73 mph. A construction sign warns that the speed limit will drop to 55 mph in 0.50 mi.
What constant acceleration (in m/s2) will bring Haley to this lower speed in the distance available?

Answer 1

Answer: $a= -1029.97 m/s^{2}$

Explanation:

We know Haley is driving with an initial velocity of $V_{o}=73mi/h$ and then she has to change to a velocity of $V_{f}=55mi/h$ in a distance $d=0.50 mi$, and we also know we are dealing with constant acceleration  $a$.

Therefore, the following equation will be useful:

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (1)

Clearing $a$:

$a=\frac{{V_{f}}^{2}-{V_{o}}^{2}}{2d}$ (2)

$a=\frac{{(55mi/h)}^{2}-{(73mi/h)}^{2}}{2(0.50 mi)}$ (3)

$a=-2304 mi/h^{2}$ (4)

Knowing $1h=3600 s$ and $1mi=1609.34 m$:

$a= -1029.97 m/s^{2}$ this is the constant acceleration that will bring Haley to lower speed