A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) At what rate is his distance from second base decreasing when he is halfway to first base? (Round your answer to one decimal place.) ft/s (b) At what rate is his distance from third base increasing at the same moment? (Round your answer to one decimal place.)

Step-by-step explanation:

Given that :

the side of the square = 90ft

The speed of the runner = 31 ft/sec

By the time the runner is halfway to the first base;  the distance covered by the runner in time(t)  is (31 t) ft and the distance half the base = 90/2 = 45 ft

Thus; 31 t = 45

t = 45/31

From the second base ; the distance is given as:

P² = (90)² + (90 - 31t )²  

P =

By differentiation with time;

At t = 45/31

= - 13.86 ft/sec

Hence, we can conclude that  as soon as the runner  is halfway to the first base, the distance to the second base is therefore decreasing by 13.86 ft/sec

b) The distance from third base can be expressed by the relation:

q² = (31t)² + (90)²

By differentiation with respect to time:

At t = 45/31

Thus, the rate at which the runner's distance is from the third base is increasing at the same moment of 13.86 ft/sec. So therefore; he is moving away from the third base at the same speed to the first base)

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse