While curling, you push a rock for 1.70 m and release it when it has a speed of 2.40 m/s. It continues to slide at constant speed for 1.40 s and then hits a rough patch of ice. It finally comes to rest 7.20 m from where it was released.
What was the curling rock's magnitude of acceleration after it hit the patch of rough ice?
To find the curling rock's magnitude of acceleration after it hit the rough patch of ice, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
- v is the final velocity (0 m/s, since it comes to rest)
- u is the initial velocity (2.40 m/s)
- a is the acceleration
- s is the displacement (7.20 m)
Substituting the known values:
0^2 = (2.40)^2 + 2a(7.20)
Simplifying:
0 = 5.76 + 14.4a
Rearranging the equation:
14.4a = -5.76
a = -5.76 / 14.4
a ≈ -0.4 m/s²
So, the magnitude of the curling rock's acceleration after hitting the patch of rough ice is approximately 0.4 m/s².