Question

What acceleration is needed to bring a car that is initially going 65.0 mi/hr to a complete stop in a distance of 57.0 m?

Answer 1

7.41 m/s^2

Step-by-step explanation

When an object is traveling in a straight line with an increase or decrease in velocity at equal intervals of time, then the object is said to be in uniform acceleration

to find the acceleration we can use the formulas

`\begin{gathered} x=v_it+(1)/(2)at^2 \\ and \\ v_f=v_1+at \end{gathered}`

so

Step 1

a)

let

`\begin{gathered} initial\text{ velocity}=v_i=65(mi)/(h) \\ final\text{ velocity= 0 \lparen complete stop-rest\rparen} \\ distance=57\text{ m=0.0354182 miles} \end{gathered}`

b) replace in equation (2)

`\begin{gathered} v_(f)=v_(1)+at \\ 0=65+at \\ -65=at \\ a=-(65)/(t) \end{gathered}`

replace in equation (1)

`\begin{gathered} x=v_(i)t+(1)/(2)at^(2) \\ 0.0354182=65t+(1)/(2)(-(65)/(t))t^2 \\ 0.0354182=65t-(1)/(2)(\frac{65}{})t^ \\ 0.0354182=65t-(65)/(2)t \\ 0.0354182=t(65-(65)/(2)) \\ 0.0354182=t(32.5) \\ t=(0.0354182)/(32.5)=0.001089\text{ hours} \end{gathered}`

now, replace in the previous equation to find a

`\begin{gathered} a=-(65)/(t) \\ a=-\frac{65(mi)/(h)}{0.001089\text{ hours }} \\ a=-59687\text{ }(miles)/(hour^2) \end{gathered}`

so, the answer is

`a=-59687.78(m\imaginaryI les)/(hour^(2))`

finally, let's convert the acceleration from miles per squared hour into meters per square second

remember that:

`\begin{gathered} 1\text{ hour = 3600 seconds } \\ 1\text{ miles=1609.34 meters} \end{gathered}`

so

`\begin{gathered} a=-59,687(m\imaginaryI les)/(hour^(2))(\frac{1\text{ hour}}{3600\text{ s}})^2(\frac{1609.34\text{ m}}{1\text{ mile}}) \\ a=-59,687\frac{\text{m}\mathrm{i}\text{les}}{\text{hour\textasciicircum{\text{2}}}}*\frac{1\text{ hour}^2}{12960000\text{s}^2}*(\frac{1609.34\text{ m}}{1\text{ mile}}) \\ a=-7.41\text{ }(m)/(s^2) \end{gathered}`

hence

the acceleration is -7.41 m/s^2

I hope this helps you