# phys

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Consider an antelope being chased by a tiger. The antelope is running at a velocity of 5.21 m/s at an angle 14.3 degree E of S. The tiger is running at a velocity of 11.64 m/s at an angle of 31.4 degree W of S. (Use East as positive x and North as postive y)

1)What are the x and y components of the velocity of the tiger with respect to the antelope?

2) What is the magnitude of the velocity of the tiger with respect to the antelope?

3) If the tiger and the antelope are 127 m apart, how long will it take the tiger to catch the antelope?

• phys -

"The antelope is running at a velocity of 5.21 m/s at an angle 14.3 degree E of S."
Antelope velocity:
θ=14.3°E of S = 270+14.3=284.3°
speed=5.21 m/s
Velocity of Antelope, Va=(5.21 m/s, 284.3°)
a.
Vax=5.21cos(284.3) m/s
Vay=5.21sin(284.3) m/s

"The tiger is running at a velocity of 11.64 m/s at an angle of 31.4 degree W of S."
θ=270°-31.4=238.6°
speed=11.64 m/s
Velocity of tiger, Vt=(11.64 m/s, 238.6°)
a.
Vtx=11.64cos(238.6) m/s
Vty=11.64sin(238.6) m/s

b.
To find the relative velocity Vr,of the tiger relative to the antelope, subtract the components of the velocity of the antelope from the velocit of the tiger:
Vr=(Vax-Vtx, Vay-Vty)
I get about (-7 m/s,-5 m/s)
Calculate the angle from the given components. I get 213.61°
c.
Magnitude of velocity, V
=√((Vtx-Vax)²+(Vty-Vay)²)

Time to catch-up, assuming the original positions are aligned to the direction Vr
= 127 m / V seconds
= 23 sec. approx.

• phys -

well, the thing is, those are the components in relation to the ground, not with each other. I can't figure out how to relate it to each other.

• phys -

b.
To find the relative velocity Vr,of the tiger relative to the antelope, subtract the components of the velocity of the antelope from the velocit of the tiger:
Vr=(Vax-Vtx, Vay-Vty)
I get about (-7 m/s,-5 m/s)
Calculate the angle from the given components. I get 213.61°

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