An lsu football defender is 6 m behind a georgia tech receiver who has the football and is d meters from making a touchdown. if the lsu player is running at 5 m/s and the receiver is running at 4 m/s, what is the furthest distance, d, that the tech player can be from the goal line and still make a touchdown?

furthest distance d = 24 m

Explanation:

LSU football defender = 6 m

LSU player is running =  5 m/s

receiver running =  4 m/s

to find out

furthest distance d

solution'we take here 2 equation

distance = velocity × time

d = 4 × t      1

6+ d = 5 × t   2

so now put d value in equation 2 we get time

6+ 4t = 5 × t  

t = 6 sec

now this t value put in equation 1

d = 4 × t  

d = 4 (6)

distance = 24 m

so furthest distance d = 24 m

total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

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potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .