A hunter with his car is chasing a deer. The car moves at 72 km/h and the deer run at speeds of
64.8 km/h. When the distance between the car and the deer is 2012 meters, the hunter fired his
shotgun. Bullets out of the gun at 200 m/s. Determine the time interval of the deer getting shot.
64.8 km/h. When the distance between the car and the deer is 2012 meters, the hunter fired his
shotgun. Bullets out of the gun at 200 m/s. Determine the time interval of the deer getting shot.
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Answer:
v_i=0\:\rm m/sv
i
=0m/s
v_f=72\:\rm km/hr=20\:\ m/sv
f
=72km/hr=20 m/s
d=45\:\rm md=45m
t=3.5\:\rm st=3.5s
(a) the cheetah’s average acceleration
a=\frac{v_f^2-v_i^2}{2d}=\rm\frac{20^2-0^2}{2\times 45}=4.44\: m/s^2a=
2d
v
f
2
−v
i
2
=
2×45
20
2
−0
2
=4.44m/s
2
(b) the cheetah’s displacement at time tt
d=\frac{at^2}{2}=\rm \frac{4.44\times 3.5^2}{2}=27.2\: md=
2
at
2
=
2
4.44×3.5
2
=27.2m
Explanation: