# Jack and Jill ran down the hill. Both started from rest and accelerated at a constant rate. Jack accelerated at .25 m/s^2 and Jill acclerated at .30 m/s^2. After running 20s, Jill fell down. a....

Jack and Jill ran down the hill. Both started from rest and accelerated at a constant rate. Jack accelerated at .25 m/s^2 and Jill acclerated at .30 m/s^2. After running 20s, Jill fell down.

a. How far did Jill get before she fell down?

b. How far had Jack traveled when Jill fell?

c. How fast was Jack running when Jill fell?

d. How long ( to the nearest second) was it after Jill fell that Jack ran ran into her and broke his crown?

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### 1 Answer

The equation of motion with constant acceleration a is

`d = v_0t + (at^2)/2` , where d is distance, t is time and `v_0` is the initial velocity. Since both Jack and Jill start from rest, their initial velocity is 0.

a) The distance Jill ran before she fell down, after 20 s, is

`d = (.3 * 20^2)/2 = 60 ` meters.

b) The distance Jack ran before Jill fell, after 20 s, is

`d = (.25*20^2)/2 = 50 ` meters

c) The speed with which Jack was running when Jill fell is

`V = at = 0.25*20 = 5 ` m/s

d) The time it would take Jack to reach Jill is the time it would take him to run remaiming 10 meters. It is the time it would take him to run 60 meters minus the time it took him to run 50 meters (20 seconds).

The time it would take him to run 60 meters is determined by

`d = (at^2)/2`

`60 = (0.25*t^2)/2`

`t^2 = (2*60)/0.25 = 480 s^2`

`t= 22` s, to the nearest second.

22 s - 20 s = 2s, so it would take Jack 2 seconds to run into Jill after she fell.

**a) 60 meters**

**b) 50 meters**

**c) 5 m/s**

**d) 2 seconds**