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Constant Acceleration of a cheetah

  • Thread starter 12345ME
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  • #1
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Homework Statement



A cheetah is hunting. Its prey runs for 3.2 s at a constant velocity of +10.0 m/s. Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time?

Homework Equations



V=Vo + a(t)

The Attempt at a Solution



I plug in the numbers to get 3.1 m/s but i know im missing something. Help with answer would be appreciated!
 

Answers and Replies

  • #2
rl.bhat
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What is the distance traveled by the cheetah?
Use the kinematic equation d = vo*t + 1/2*a*t^2 to find a.
 
  • #3
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this is all the problem says, doesnt give a distance.
 
  • #4
rl.bhat
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this is all the problem says, doesnt give a distance.
Velocity of the cheetah is given, time taken be it is given. What is the relation between distance, velocity and time?
 
  • #5
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A cheetah is hunting. Its prey runs for 3.2 s at a constant velocity of +10.0 m/s.
What distance does the prey runs?

Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time?
You know the d now, t is 3.2s ... find a
 
  • #6
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still have no idea what the distance is
 
  • #7
rl.bhat
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still have no idea what the distance is
Open any text book and find the relation between velocity, distance and time.
 
  • #8
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distance= velocity x time right?
 
  • #9
rl.bhat
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  • #10
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but when you use d = vo*t + 1/2*a*t^2

i get 0= 1/2*a*3.2^2

which equals 0
 
  • #11
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lost
 
  • #12
rl.bhat
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  • #13
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d = 3.2 s*10 m/s
ya but so does Vo*t on the right side of the equation so they cancel each other out
 
  • #14
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it comes out to
32=32 + 1/2*a*3.2^2
0=1/2*a*3.2^2
?
 
  • #15
rl.bhat
Homework Helper
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it comes out to
32=32 + 1/2*a*3.2^2
0=1/2*a*3.2^2
?
Initial velocity of cheetah is zero.
 
  • #16
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6.25 thank you!
 

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