Question 1162207: One fine winter morning, you are walking your dog,Tootsie, in the park. Tootsie sees a cat. For the first 10.0 seconds, you hold the leash as tight as you can, and you manage to keep Tootsie in one spot. Then, Tootsie gets loose. He accelerates at 4.21 (m/s)/s. He reaches his maximum speed of 421.7 m/min. Then, Tootsie collides with the cat. The collision causes him to slow down at a rate of 3.50 (m/s)/s until he finally stops. Calculate the distance travelled by Tootsie from the time he saw the cat to the time he finally stopped. (Assume acceleration is constant.)
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! One fine winter morning, you are walking your dog,Tootsie, in the park. Tootsie sees a cat. For the first 10.0 seconds, you hold the leash as tight as you can, and you manage to keep Tootsie in one spot. Then, Tootsie gets loose. He accelerates at 4.21 (m/s)/s. He reaches his maximum speed of 421.7 m/min. Then, Tootsie collides with the cat. The collision causes him to slow down at a rate of 3.50 (m/s)/s until he finally stops. Calculate the distance travelled by Tootsie from the time he saw the cat to the time he finally stopped. (Assume acceleration is constant.)
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421.7 m/min = 7.028 m/sec
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The first 10 seconds is not relevant as he's stationary.
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v = a*t
t = v/a = 7.028/4.21 = 1.669 seconds of acceleration
His average speed for the 1.669 seconds is 7.028/2
distance = 1.669*7.028/2 = ~5.8667 meters until his collision with the cat.
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The time until he's stopped = 7.028/3.5 seconds = 2.008 seconds
The average speed after the collision = 7.028/2
Distance after the collision = 2.008*7.028/2 = 7.056 meters
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7.056 + 5.8667 = 12.923 meters
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PS Tootsie is not a good name for a male dog.
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