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Mathematics 9 Online
OpenStudy (cruffo):

Bob is 70 miles away from home. He starts his trip home going 70 mph, and slows down continuously. (a) He plans to stop at the store one mile away from home. How long will it take him to get to the store? (b) How long will it take him to get from the store to his home if he uses the same driving strategy?

OpenStudy (anonymous):

V=70-at I assume at he slows down continuously and stopping at his home. Let's use a kinematic equation vf^2=vi^2+2ad 0 =70-2a(70) 70=-140a a=-.5 mile per hour^2

OpenStudy (anonymous):

Stopping at the store x=x0+v0t+1/2 at^2 60=0+70+1/2(-.5)t^2 Find t

OpenStudy (anonymous):

at what rate is he slowing down continuously? is he losing 1 m/h every minute? 5 m/h? 10?

OpenStudy (anonymous):

I already found that, it is -0.5

OpenStudy (cruffo):

Continuously - if he is d miles away, then he is traveling at d miles per hour.

OpenStudy (anonymous):

I am confident that I did it right

OpenStudy (cruffo):

I was thinking the solution involved an integral. Let T be the total time, then \[T = \int_1^{70} f(t) dt\] question is what's f(t) ?

OpenStudy (cruffo):

Assuming linear speed: d = rt then t = d/r. so I think f(t) = distance/rate.

OpenStudy (anonymous):

Well we know that it is constant accelerationV=\[\int\limits -a \, dt\], to find velocity we integrate we get =-at+V0 To get position we integrate again X=\[\int\limits -\text{at}+V\text{ }\text{dt}\] We get -1/2 at^2+Vt+X0

OpenStudy (cruffo):

Wouldn't this give the total change in position?

OpenStudy (anonymous):

Yes, but integration find us the forumla which we can use to solve problem like this

OpenStudy (cruffo):

hum... the solution to part a is ln(70).

OpenStudy (cruffo):

I think what's happening is the following. D = RT He is going d mph when he is d miles from home. How long does it take him to go the first mile when he is going 70mph? 1/70 h Of course this changes continuously, but the idea is that it will always take him 1/d h to go 1 mile when he is d miles from home. Thus t = 1/d so d = 1/t. And let the change in time be dt. so we get \[\int_1^{70} \frac{dt}{t}\] Which would get me to the solution.

OpenStudy (cruffo):

May be someone can explain this a little better.

OpenStudy (cruffo):

imranmeah91 -Since velocity is changing continuously, we definitely have acceleration. But how would you know that acceleration is a constant, and not some other function of time?

OpenStudy (anonymous):

I assumed that acceleration is constant

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