Tremont the Turkey finds out that he\'s scheduled to be someone\'s Thanksgiving

Question

Tremont the Turkey finds out that he's scheduled to be someone's Thanksgiving feast this year, so he decides to run away. After launching himself over the fence, he begins accelerating as he runs across the field toward the woods. Fortunately for him, he is able to maintain a constant acceleration across the length of the 75 yard field.

Five seconds after he launches over the fence, Bertie Bulldog, who is patrolling the edge of the woods, realizes that Tremont is out of the enclosure, and begins running to intercept him at the edge of the field. Bertie has an acceleration of 4.0 ft/s^2 for the first 3.0 seconds of his run, and after that he maintains constant speed. To reach the point where Tremont will enter the woods, Bertie must travel 110 yards.

If Tremont escapes by the skin of his beak (he reaches the woods just as Bertie reaches him), how fast was Tremont going when he reached the woods?

Explanation / Answer

d1 = 75 yards = 225 ft

suppose turkey's acc is a.

then speed after travelling 225 ft.

vf^2 - vi^2 = 2 a d

v^2 - 0^2 = 2 a 225

v = 15 sqrt(2a) ....after travelling 75 ft.

time taken to travel 75 ft.

d = v0 t + a t^2 /2

75 = 0 + a t^2 /2

t = sqrt(150 / a)

for bulldog:

distance travelled in first 3 sec.

d = 0 + (4 x 3^2 /2 ) = 18 ft  

v = (4 x 3) = 12 ft/s

after travels, 110 - 18 = 92 ft with 12 ft/s

t = 92/12 =7.67 sec

total time turkey travelled = 5 + 3 + 7.67 =15.67 sec

it travels 110 -75 = 35 ft wth constant 15 sqrt(2a) speed in (15.67 -sqrt(150 / a) ) sec.

so 15 sqrt(2a) ( 15.67 -sqrt(150 / a) ) = 35

a = 0.787 m/s^2

speed of turkey = 15 sqrt(2 x 0.787)

= 18.8 ft/s

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