Abaseball diamond is a square with sides of length 90 ft. a batter hits the ball and runs toward first base with a speed of 24 ft/s.

at what rate is his distance from second base changing when he is halfway to first base?

- 10.7 m/s

Explanation:

The distance between the batter and the second base is the length of the diagonal formed by him, his distance from first base, and the distance between first and second base.

Use these variables:

distance from the batter to second base: ddistance from the batter to first base: xdistance between first and second base: 90 ft (constant).

Using Pythagora's theorem:

d² = x² + 90²

Knowing that the speed of the bater toward first base is 24 ft/s, his distance, x, can be written as: x = 90 - 24t

And the equation for d becomes:

d² = 90² + (90 - 24t)²

You want to find the rate of change of d. Call it d' and use implicit differentiation:

2d.d' = 2(90-24t) × (-24)

Solve for d':

d' = -48(90 - 24t) / (2d)d' = -24 (90 - 24t) / d

You can determine t and d from the fact that the batter is halfway to first base:

For the time, use the fact that he has run 45 ft and the spedd is 24 ft/s. So, the time = distance / speed = 45m / 24 m/s

Substitute in the equation for d':

That is d' = - 10.7 m/s

The negative sign means that his distance from second base is decreasing (he is closer from second base as he is coming closer to first base).

At a constant it will take one hour to travel 32 mi.at that speed it will take 1/2 hr to travel the remaining 16 mi.it will take 1+1/2 hr to travel the 48mi @ 32mph.so 1 hr = 60 min& 1/2 hr = 30 min1+1/2 hr = 90 min.there are 60s in each 90 × 60 = 3600 seconds.the real question seriously is he why are they driving so slow? ! lol.

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