A softball player is running at 4.88 m/sec when she slides into second base coming to a stop in .872 seconds. How far did she slide, and what was her acceleration?​

d=v1t - .5at^2

d=4.88 x .872 - 0.5 x (4.88/0.872) x 0.872^2

d=4.255 - 2.12

d= 2.135m

Explanation:

acceleration is negative because she is slowing down.

(75°f − 32) × 5/9 = 23.889°c

i believe the answer is a hope it and is correct