- 4

- 0

1. The problem statement, all variables and given/known data

A cheetah is estimated to be able to run at a maximum speed of [tex]100km.h^1[/tex] whilst and antelope can run at a maximum speed of [tex]65km.h^1[/tex]. A cheetah at rest sees an antelope at rest and starts running towards it. The antelope immediately starts running away. Both cheetah and antelope are assumed to move with constant acceleration and reach their maximum speeds in 4 seconds. Assuming that both run along the same straight line and that the cheetah catches the antelope in 15 seconds, find the distance between the animals when they first started moving.

[tex]v=u+at[/tex]

[tex]s=ut+1/2at^2[/tex]

[tex]v^2=u^2+2as[/tex]

[tex]s=1/2(u+v)t[/tex]

I have been staring at this for a good hour now and am yet to come up with the answer which I know to be 126m. How do i find this value using those equations and the data given?

here is what i have gotten so far but the final answer is not 126m which i know to be correct, any ideas?

cheetah-[tex]v=250/9[/tex][tex]u=0[/tex] [tex]t=4[/tex] [tex]a=?[/tex]

[tex]250/9 =4a[/tex]

[tex](250/9)/4=125/18[/tex]

[tex]125/18=a[/tex]

antelope-[tex]v=325/18[/tex] [tex]u=0[/tex] [tex]t=4[/tex] [tex]a=?[/tex]

[tex](325/18)/4=a[/tex]

[tex]325/72=a[/tex]

cheetah displacement -

[tex]s=ut+1/2at^2[/tex]

[tex]s=0x4+1/2x125/18x4^2[/tex]

[tex]s=500/9[/tex]

[tex]s=1/2(u+v)t[/tex]

[tex]1/2(0+250/9)11[/tex]

[tex]s=1375/9[/tex]

antelope displacement -

[tex]s=ut+1/2at^2[/tex]

[tex]s=0x4+1/2x325/72x4^2[/tex]

[tex]s=325/9[/tex]

[tex]s=1/2(u+v)t[/tex]

[tex]1/2(0+325/18)11[/tex]

[tex]s=3575/36[/tex]

[tex]s=(625/3)-(1625/12)=875/12[/tex]

Does that look correct? According to the answer i know to be correct (its on the back of the assignment paper) it is wrong :s

Thanks again for this!

A cheetah is estimated to be able to run at a maximum speed of [tex]100km.h^1[/tex] whilst and antelope can run at a maximum speed of [tex]65km.h^1[/tex]. A cheetah at rest sees an antelope at rest and starts running towards it. The antelope immediately starts running away. Both cheetah and antelope are assumed to move with constant acceleration and reach their maximum speeds in 4 seconds. Assuming that both run along the same straight line and that the cheetah catches the antelope in 15 seconds, find the distance between the animals when they first started moving.

**2. Equations I have been given**[tex]v=u+at[/tex]

[tex]s=ut+1/2at^2[/tex]

[tex]v^2=u^2+2as[/tex]

[tex]s=1/2(u+v)t[/tex]

I have been staring at this for a good hour now and am yet to come up with the answer which I know to be 126m. How do i find this value using those equations and the data given?

here is what i have gotten so far but the final answer is not 126m which i know to be correct, any ideas?

cheetah-[tex]v=250/9[/tex][tex]u=0[/tex] [tex]t=4[/tex] [tex]a=?[/tex]

[tex]250/9 =4a[/tex]

[tex](250/9)/4=125/18[/tex]

[tex]125/18=a[/tex]

antelope-[tex]v=325/18[/tex] [tex]u=0[/tex] [tex]t=4[/tex] [tex]a=?[/tex]

[tex](325/18)/4=a[/tex]

[tex]325/72=a[/tex]

cheetah displacement -

**first 4 secs**[tex]s=ut+1/2at^2[/tex]

[tex]s=0x4+1/2x125/18x4^2[/tex]

[tex]s=500/9[/tex]

**final 11 secs**[tex]s=1/2(u+v)t[/tex]

[tex]1/2(0+250/9)11[/tex]

[tex]s=1375/9[/tex]

**total S**= [tex]625/3[/tex]antelope displacement -

**first 4 secs**[tex]s=ut+1/2at^2[/tex]

[tex]s=0x4+1/2x325/72x4^2[/tex]

[tex]s=325/9[/tex]

**final 11 secs**[tex]s=1/2(u+v)t[/tex]

[tex]1/2(0+325/18)11[/tex]

[tex]s=3575/36[/tex]

**total S**= [tex]1625/12[/tex][tex]s=(625/3)-(1625/12)=875/12[/tex]

Does that look correct? According to the answer i know to be correct (its on the back of the assignment paper) it is wrong :s

Thanks again for this!

Last edited: