Calculating Headstart for a Cheetah Chasing an Antelope

[aydemir]

1. The Cheetah is the fastest land animal in the world. It can Accelerate from rest to 20m/s in 2 second, and has a top speed of 30 m/s. it can only maintain its top speed for 450 metres before it has to stop and rest. in contrast an antelope runs at a top speed of 22 m/s for much longer periods.

(cheetah and antelope have the same acceleration)

if the cheetah chases the antelope, both starting from rest, what is the maxiumum headstart the cheetah can give the antelope.



2. The Equations of Motion e.g. "v=u+at" ect.



3. i worked out that the acceleration of the cheetah and antelope to be 10 m/s (squared)
by simply using "acceleration = change in Velocity / time" which was
"(20-0)/2 = 10 m/s"

acceleration distance and time for the cheetah is 45m to reach top speed OR 3s to
reach top speed because "v+u+at therefore 30=0 + 10 * t (t=3s)
and x+ut+1/2 at (squared) (x= 45m)"

from further simple calculations i got the time for the cheetah to run 450m which was
16.5 seconds "x+450 - 45 = 405" 405m is the distance left over after the cheetah is
done accelerating, then i used "x=ut+1/2 at(squared)" (note that the a=0 so its really
x=ut)

the distance it takes the antelope to reach top speed is 24.2m and the time is 2.2s

now this is where I am stuck, so if anyone could help i would be extremely happy, thankyou
very much

 

[xcvxcvvc]

1. The Cheetah is the fastest land animal in the world. It can Accelerate from rest to 20m/s in 2 second, and has a top speed of 30 m/s. it can only maintain its top speed for 450 metres before it has to stop and rest. in contrast an antelope runs at a top speed of 22 m/s for much longer periods.

(cheetah and antelope have the same acceleration)

if the cheetah chases the antelope, both starting from rest, what is the maxiumum headstart the cheetah can give the antelope.
2. The Equations of Motion e.g. "v=u+at" ect.
3. i worked out that the acceleration of the cheetah and antelope to be 10 m/s (squared)
by simply using "acceleration = change in Velocity / time" which was
"(20-0)/2 = 10 m/s"

acceleration distance and time for the cheetah is 45m to reach top speed OR 3s to
reach top speed because "v+u+at therefore 30=0 + 10 * t (t=3s)
and x+ut+1/2 at (squared) (x= 45m)"

from further simple calculations i got the time for the cheetah to run 450m which was
16.5 seconds "x+450 - 45 = 405" 405m is the distance left over after the cheetah is
done accelerating, then i used "x=ut+1/2 at(squared)" (note that the a=0 so its really
x=ut)

the distance it takes the antelope to reach top speed is 24.2m and the time is 2.2s

now this is where I am stuck, so if anyone could help i would be extremely happy, thankyou
very much

Everything you did so far looks great and is what I would have done. Next, I'd calculate how much distance the antelope can run in that same time, so use v * t where t is 16.5 - 2.2 to find distance ran with constant velocity. Then add the 24.2m you calculated earlier. This distance will be lower than 450. If you subtract it from 450, that is the maximum headstart allowed. Intuitively, this makes sense, but you can see it mathematically too:
[tex] R(t) = \frac{1}{2} a t^2 + v_o * t + R_o[/tex]
For us, however, acceleration and constant velocity are different times:
[tex] R(t) = \frac{1}{2} a t_1^2 + v_o * t_2 + R_o[/tex]
Solve for R_o, the initial distance, so that the antelope will run a distance R(t) equal to the cheetah (450 M)
[tex] R_o = R(t) - \frac{1}{2} a t_1^2 - v_o * t_2[/tex]
[tex] R_o = 450 - \frac{1}{2} 10*2.2^2 - 22 * (16.5 - 2.2)[/tex]

 

[aydemir]

thankyou very much bro, i get it now :)