Answer :
30 mph
Explanation
Step 1
Draw the situation
to solve this we need to use the expession:
[tex]\begin{gathered} v_f=v_1-at \\ \text{whre v}_fis\text{ the final velocity} \\ vi\text{ is the initial velocity} \\ a\text{ is the acceleration} \\ t\text{ is the time} \end{gathered}[/tex]so
a) Set the equations
so
[tex]\begin{gathered} v_f=v \\ v_i=50\text{ mph} \\ \text{time}=0.02\text{ hours}(1.2\text{ minutes)} \\ \text{distance}=\text{ 0.8 mi} \\ \text{now, replace} \\ v_f=v_1-at \\ v=50mph-a(t_1) \\ v=50-a(0.02) \\ v=50-0.02a\rightarrow equation(1) \end{gathered}[/tex]and
[tex]\begin{gathered} v^2_f=v^2_0-2ax \\ v^2=50^{^2}-2a(0.8) \\ v^2=2500-1.6a\rightarrow equiation(1B) \end{gathered}[/tex]so, we have 2 equatonis
[tex]\begin{gathered} v=50-0.02a\rightarrow equation(1) \\ v^2=2500-1.6a\rightarrow equiation(1B) \end{gathered}[/tex]Step 2
solve the equations
a) isolate a in equation ( 1) and replace in equation (1B)
so
[tex]\begin{gathered} v=50-0.02a\rightarrow equation(1) \\ v-50=-0.02a \\ a=\frac{v-50}{-0.02} \\ \text{replace ine (1B)} \\ v^2=2500-1.6a\rightarrow equiation(1B) \\ v^2=2500-1.6(\frac{v-50}{-0.02})\rightarrow equiation(1B) \\ v^2=2500+80(\frac{v-50}{\square}) \\ v^2=2500+80v-4000 \\ v^2-80v+1500=0 \\ v=30 \end{gathered}[/tex]