What's the value of the two#?# ? I don't understand what's the question is asking actually and have no idea on what method to use to calculate. :(

You are testing your brand new Ferrari Testarossa. To see how well the brakes work you accelerate to 100 miles per hour, slam on the brakes, and determine that you brought the car to a stop over a distance of 487 feet. Assuming a constant deceleration you figure that that deceleration is #?# feet per second squared. (Enter a positive number.)

I trust that you don't have the courage to try this, but that night you wonder how long it would take you to stop (with the same constant deceleration) if you were moving at 200 miles per hour. Your stopping distance would be #?# feet. (Enter a number, not an arithmetic expression.)

1 Answer
Sean
Nov 28, 2017

Part one:

#a=-146.67/6.64=22.08# #ft# / #s^2#

Part two:

#d=146.67(13.29)=1949.24# feet

Explanation:

.

#v=v_i+at#

where #v# is your final velocity and #v_i# is your initial velocity and #a# is acceleration.

In this problem:

#v=0# and #v_i=100# miles per hour. We have to make sure we use all the same units. That means we have to convert the initial velocity to feet per second.

Each mile is 5280 feet. That says that the car's velocity was:

#5280*100=528000# feet per hour. Now, we have to figure out what it was per second. So, we divide that by #3600# because there are #3600# seconds in one hour:

#v_i=528000/3600=146.67# #ft# / #s#

Now, we can plug the value into the velocity equation above to solve for acceleration:

#0=146.67+at#

#-146.67=at#

#a=-146.67/t#

It is negative because we have deceleration.

The distance formula is:

#d=v_it+1/2at^2# which is the integral of the velocity function above.

in our problem #d=487#. Let's plug that and #a# in and solve for #t#:

#487=146.67t+1/2(-146.67/t)t^2#

#487=146.67t-73.34t#

#487=73.34t#

#t=487/73.34=6.64# seconds is the time it takes for the car to come to a stop.

Now, we can plug this into the acceleration equation and find #a#:

#a=-146.67/6.64=22.08# #ft# / #s^2#

For part two, #v_i=200# miles per hour which is #293.34# feet per second. Then:

#a=-293.34/t#

Since we have the same constant deceleration, we can plug it in and find the time it takes for the car to stop from #200# miler per hour:

#22.08=-293.34/t#

#22.08t=-293.34#

#t=-293.34/22.08=13.29# seconds

#d=293.34t+1/2(-293.34/t)t^2#

#d=293.34t-146.67t#

#d=146.67t#

#d=146.67(13.29)=1949.24# feet

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