1) A car traveling at a constant velocity of 60 miles/hour must come to a stop i

Question

1)
A car traveling at a constant velocity of 60 miles/hour must come to a stop in 150 feet using a constant rate of deceleration. What is the rate of deceleration that must be applied? Note: 1 mile = 5280 feet & 1 hour = 3600 seconds

answer: -25.8ft/s^2


2)A cat runs past a dog that is standing still. The cat is running at a constant speed 12 feet per second. At the instant the cat passes the dog, the dog starts running with a constant acceleration of 6ft/s

a. When will the dog catch the cat? b. How far will the dog have to run?


answer:
a)4s
b)48 feet


PLEASE SHOW WORK STEP BY STEP!

THANKS:)

1)
A car traveling at a constant velocity of 60 miles/hour must come to a stop in 150 feet using a constant rate of deceleration. What is the rate of deceleration that must be applied? Note: 1 mile = 5280 feet & 1 hour = 3600 seconds

answer: -25.8ft/s^2


2)A cat runs past a dog that is standing still. The cat is running at a constant speed 12 feet per second. At the instant the cat passes the dog, the dog starts running with a constant acceleration of 6ft/s

a. When will the dog catch the cat? b. How far will the dog have to run?


answer:
a)4s
b)48 feet


PLEASE SHOW WORK STEP BY STEP!

THANKS:)

12 feet per second. At the instant the cat passes the dog, the dog starts running with a constant acceleration of 6ft/s

a. When will the dog catch the cat? b. How far will the dog have to run?


answer:
a)4s
b)48 feet


PLEASE SHOW WORK STEP BY STEP!

THANKS:)
a. When will the dog catch the cat? b. How far will the dog have to run?

Explanation / Answer

v=60m/h=88 ft/s

s=150ft

v^2=-2as

-a=v^2/2s=25.8ft/s^2

2)

cat speed vc=12ft/s

dog acceleration ad=6ft/s^2

a)t=2*12/6=4s

b)

s=1/2at^2

s=1/2*6*4^2=48ft

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