Constant Acceleration of a cheetah

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Homework Help Overview

The problem involves a cheetah attempting to match the distance covered by its prey, which runs at a constant velocity. The cheetah starts from rest and must maintain a constant acceleration to cover the same distance in the same time frame.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between distance, velocity, and time, questioning how to derive the distance traveled by the prey. There are attempts to apply kinematic equations, but confusion arises regarding the initial conditions and the calculations involved.

Discussion Status

The discussion reflects a mix of understanding and uncertainty, with some participants providing guidance on the relationship between distance, velocity, and time. However, there is no explicit consensus on the correct approach to find the distance or the acceleration needed.

Contextual Notes

Participants note that the problem does not explicitly provide the distance, leading to confusion. There are references to textbook knowledge and the need to clarify the relationship between the variables involved.

12345ME
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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 I am missing something. Help with answer would be appreciated!
 
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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.
 
this is all the problem says, doesn't give a distance.
 
12345ME said:
this is all the problem says, doesn't give a distance.
Velocity of the cheetah is given, time taken be it is given. What is the relation between distance, velocity and time?
 
12345ME said:
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
 
still have no idea what the distance is
 
12345ME said:
still have no idea what the distance is
Open any textbook and find the relation between velocity, distance and time.
 
distance= velocity x time right?
 
12345ME said:
distance= velocity x time right?
Right.
 
  • #10
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
lost
 
  • #12
12345ME said:
lost

d = 3.2 s*10 m/s
 
  • #13
rl.bhat said:
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
it comes out to
32=32 + 1/2*a*3.2^2
0=1/2*a*3.2^2
?
 
  • #15
12345ME said:
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
6.25 thank you!
 

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