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## Homework Statement

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/s

^{2}.

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?

## Homework Equations

S=vt

S=1/2 at

^{2}

## The Attempt at a Solution

1. Solution for the first question.

Function for the man displacement is: S

_{man}=4t+6

Function for the bus displacement is: S

_{bus}=1/2(1.2)t

^{2}=0.6t

^{2}

When the man catches the bus: S

_{man}=S

_{bus}4t+6 = 0.6t

^{2}0.6t

^{2}-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: S

_{man}=4t+10

S

_{man}=S

_{bus}0.6t

^{2}-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 ?