A man runs at a speed of 4.0 m/s to catch a standing bus. When his distance to the door is 6.0 m, the bus starts moving forward and continues with a constant acceleration of 1.2 m/s2.
1. How many seconds does it take for the man to reach the door ?
2. If at the beginning he is 10.0 m from the door, will he (running at the same speed) ever catch up?
The Attempt at a Solution
1. Solution for the first question.
Function for the man displacement is: Sman=4t+6
Function for the bus displacement is: Sbus=1/2(1.2)t2=0.6t2
When the man catches the bus: Sman=Sbus 4t+6 = 0.6t2 0.6t2-4t-6 = 0 (t>0) → t≈ 7.9 s.
2.Solution for the second question.
If at the beginning he is 10.0 m from the door, function for his displacement is: Sman=4t+10
Sman=Sbus 0.6t2-4t-10 = 0 (t>0) → t ≈ 8.6 s
Therefore if at the beginning he is 10 m from the door, he will catch the bus after 8.6 s.
I wonder if my solutions are correct? because I'm not sure whether I should add 6 meters and 10 meters or subtract them from the displacement function of the man ?