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:

0.29$\text{m}\text{s}^{2}$

Explanation:

Given: 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.

To Find: constant acceleration (in $\text{m}\text{s}^{2}$) that will bring Haley to this lower speed in the distance available

Solution:

we know that,

$1$ mile=$1.609$km

$1$ km$\setminus$h = $5\setminus18 m\setminus s$

Initial Speed of Haley(u)=$73$ mph

                                     $73\times 1.609\times 5\setminus 18$

                                     $32.63$$\text{m}\setminus\text{s}$

Final Speed of Haley(v)= $55$ mph

                                     $55\times 1.609\times 5\setminus 18$

                                     $24.58$$\text{m}\setminus\text{s}$

The distance to be travelled  while lowering the speed(S)=$0.5$ mile

                                     $0.805$ km

                                     $805$ m

according to third equation of motion,

                     $\text{v}^{2}-\text{u}^{2}=2\text{a}\text{S}$

as speed is lowering down the acceleration will be in the opposite direction of motion, hence acceleration will be negative, equation will become

                    $\text{u}^{2} -\text{v}^{2}=2a\text{S}$

putting values,

                     $32.63^{2}$-$24.58^{2}$=$2a\times805$

                     a=$460.55\setminus1610$

                     a≅$0.29$$\text{m}\text{s}^{2}$

the constant acceleration that will drop speed to $55 \text{mph}$ is $0.29$$\text{m}\text{s}^{2}$

                                                                                                                                   

Answer 2

Picture is my answer