While curling, you push a rock for 2.00 m and release it when it has a speed of 1.90 m/s. It continues to slide at constant speed for 0.700 s and then hits a rough patch of ice. It finally comes to rest 8.30 m from where it was released. What was the curling rock's magnitude of acceleration after it hit the patch of rough ice?

Answer 1

vf^2=vi^2+2ad or
0=1.9^2 + 2a*8.3 solve for acceleration a.

Answer 2

initial speed = 1.90 m/s at release
slides at that speed for 0.7 s so goes 1.33 m
slows at constant acceleration
stops in 8.30 - 1.33 = 6.97 meters
average speed during stop = 1.90/2 = 3.485 m/s
so time to stop = 6.97 /3.485 = 2 seconds on the rough ice
a = change in speed / time = 1.90/2 = 0.95 m/s^2

Answer 3

slides after release at 1.9 until it hits the rough spot so only brakes for about 7 meters