Question

A car is traveling at 50mph applies the brakes and the car decelerate at a constant rate. What is the velocity of the car 0.8mi farther along the road if the car reaches this point 1.2 min after brakes are applied?

Answer 1

30 mph

Step-by-step explanation

Step 1

Draw the situation

to solve this we need to use the expession:

`\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}`

so

a) Set the equations

so

`\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}`

and

`\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}`

so, we have 2 equatonis

`\begin{gathered} v=50-0.02a\rightarrow equation(1) \\ v^2=2500-1.6a\rightarrow equiation(1B) \end{gathered}`

Step 2

solve the equations

a) isolate a in equation ( 1) and replace in equation (1B)

so

`\begin{gathered} v=50-0.02a\rightarrow equation(1) \\ v-50=-0.02a \\ a=(v-50)/(-0.02) \\ \text{replace ine (1B)} \\ v^2=2500-1.6a\rightarrow equiation(1B) \\ v^2=2500-1.6((v-50)/(-0.02))\rightarrow equiation(1B) \\ v^2=2500+80((v-50)/(\square)) \\ v^2=2500+80v-4000 \\ v^2-80v+1500=0 \\ v=30 \end{gathered}`