Question

While curling, you push a rock for 1.70 m and release it when it has a speed of 1.70 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 9.20 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

Answer on Question #62208-Physics – Mechanics | Relativity

While curling, you push a rock for $1.70\mathrm{m}$ and release it when it has a speed of $1.70\mathrm{m/s}$. It continues to slide at constant speed for 0.700 s and then hits a rough patch of ice. It finally comes to rest $9.20\mathrm{m}$ from where it was released.

What was the curling rock's magnitude of acceleration after it hit the patch of rough ice?

Solution

The distance from the place where a rock was released to final position is

$$D = d_1 + d_2 = vt_1 + vt_2 - \frac{at_2^2}{2}.$$ $$v - 0 = at_2 \rightarrow t_2 = \frac{v}{a}.$$ $$D = vt_1 + v\left(\frac{v}{a}\right) - \frac{a\left(\frac{v}{a}\right)^2}{2} = vt_1 + \frac{v^2}{2a}.$$ $$\frac{v^2}{2a} = D - vt_1$$ $$a = \frac{v^2}{2(D - vt_1)} = \frac{1.7^2}{2(9.2 - 1.7 \cdot 0.7)} = 0.180\,\frac{\mathrm{m}}{\mathrm{s}^2}.$$

Answer: $0.180\,\frac{\mathrm{m}}{\mathrm{s}^2}$.

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