in attempt to catch a bus, a women is running at a constant speed of 6 m/s. When she is 12 m away from the bus

Question

, the bus takes off with a constant acceleration of 1.20 m/s^2...

--- the position of the lady as a funtion of time is 6t & the bus is
12 + .6t^2

---the times when she catches the bus are 2.77s & 7.23s

i got those two things right but i cant get the A & B right help..

Find....

a. the place where she catches the bus.
b. find the speed of the bus when she catches the bus.

Answer 1

OK, the position of the bus is given by 12 + 0.6t^2.
The position of the lady is given by 6t.

When the lady and bus are in the same place,
12 + 0.6t^2 = 6t

Rearranging:
(3/5)t^2 - 6t + 12 = 0

Multiplying through by (5/3):
t^2 - 10t + 20 = 0

Shove this through the quadratic equation, and you get:
t = (10 +/- sqrt(100 - 80))/2
t = 5 +/- sqrt(20)/2
t = 5 +/- 2sqrt(5)/2
t = 5 +/- sqrt(5)
t = 5 +/- 2.236
t = 2.76 seconds and t = 7.24 seconds.

Obviously, the 2.76 second answer is the one that applies.

After 2.76 seconds, the speed of the bus is:
1.20 (m/s^2) * 2.76s = 3.32 m/s

Hope that helps!