Expert: Scotto - 9/23/2010

Question
I'm supposed to use the quadratic equation for the following
physics problem:

A bus has stopped to pick up riders. A student is
approaching the bus from behind on the sidewalk at a
constant velocity of 5 m/s. When he is 10 m from the bus,
the bus pulls forward with a constant acceleration of 1
m/s2. If the student continues running, what is the shortest
time (in seconds) before he catches the bus? Set the two
equations as equal and solve the quadratic.

I know that for the student:
x=10+5t
Bus:
x=.5(1)t^2

So I set up the equation:

10+5t=.5(1)t^2

This is where I'm stuck I don't how to solve for t (it's
been awhile since I took a math course).

How do you isolate t, I know that t has to =5 - square root
of 5.
I know the answer but I want to know how you get from
10+5t=.5t^2 to 5 - square root of 5??
Thanks!

Answer
Assuming the equation is right, then 10 + 5t = 0.5*t².

That is the same as 10 + 5t = t²/2.

Putting all of the variables on one side gives us 10 + 5t - t²/2 = 0.

Eliminating the fraction means to multiply by 2 and get
20 + 10t - t² = 0.  This was done just to make it easier
to deal with.

Putting the t² term out front means taking the negative of
all terms and reversing the order, given t² - 10t - 20 = 0.

The quadratic equation says that with at² + bt + c = 0, the two solutions are t = (-b+√(b²+4ac))/2a and t = (-b+√(b²-4ac))/2a.
Here, a = 1, b = -10, and c = -20.  That means that -b = 10,
b² = 100, -4ac = -4(1)(-20) = 80, and 2a = 2(1) = 2.

Solving involves using the quadratic equation and getting
t = (10+√(100+80))/2 and t = (10-√(100+80))/2.

That is the same as t = (10+√180)/2 and t = (10-√180)/2.

The 10/2 is 5 and (√180)/2 = √180/√4 = √(180/4) = √45.

That is the same as t = 5+√45 and t = 5-√45,
and these are the two possible answers.

Now √45 can be seen to be greater than 6, since √36 = 6, so the 2nd answer above is negative.  This means the answer is the 1st one.