Atrain starts from a station with a constant acceleration of at = 0.40 m/s2. a passenger arrives at the track time t = 6.0s after the end of the train left the very same point. what is the slowest constant speed at which she can run and catch the train. sketch curves for the motion of the pasenger and the train as functions of time.

4.8 m/s  

Explanation:

When she catches the train,

The formula for the distance covered by the train  is

d = ½at² = ½ × 0.40t² = 0.20t²

The passenger starts running at a constant speed 6 s later, so her formula is

d = v(t - 6.0)

The passenger and the train will have covered the same distance when she has caught it, so

(1) 0.20t² = v(t - 6.0)

The speed of the train is

v = at = 0.40t

The speed of the passenger is v.

(2) 0.40t = v

Substitute (2) into (1)

0.20t² = 0.40t(t - 6.0) = 0.40t² - 2.4 t

Subtract 0.20t² from each side

0.20t² - 2.4t = 0

Factor the quadratic

t(0.20t - 2.4) = 0

Apply the zero-product rule

t =0     0.20t - 2.4 = 0

                   0.20t = 2.4

(3)                      t = 12

We reject t = 0 s.

Substitute (3) into (2)

0.40 × 12 = v

            v = 4.8 m/s

The slowest constant speed at which she can run and catch the train is 4.8 m/s.

A plot of distance vs time shows that she will catch the train 6 s after starting. Both she and the train will have travelled 28.8 m. Her average speed is 28.8 m/6 s = 4.8 m/s.

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unbalanced force

explanation: