at t=38, one wants the separation be 1 meter.
distance(38)=speedgazelle*38-speedcheeta(38)+intialseparation
change the speeds to m/s, then
1=xxx*38-YYY*38+initialseparation
solve for separation
A cheetah can maintain its maximum speed of 120 km/hr for 38 seconds. What minimum distance must a gazelle running 74 km/hr be ahead of the cheetah to escape?
A) 417.2 m
B) 3778 m
C) 624.3 m
D) 485.6 m
E) 1748 m
3 answers
so, with that i got 484.26.
they obviously solved for separation to be 0 at 38s, which doesn't make sense because the gazelle would die.. but do you think d then is the right answer?
they obviously solved for separation to be 0 at 38s, which doesn't make sense because the gazelle would die.. but do you think d then is the right answer?
Dc = 120,000m/h * (38/3600)h = 1266.7 m. = Distance covered by cheetah.
Dg + 74,000m/h*(38/3600))h = 1266.7.
Dg = 485.6 m.
Dg + 74,000m/h*(38/3600))h = 1266.7.
Dg = 485.6 m.