Question

a ball is rolling on the horizantal travels 12m and comes to a complete stop. the ball is subject to an acceleration of -1.2m/s squared. calulate original velocity, time to stop, and average velocity

Answer 1

Question#19489

a ball is rolling on the horizontal travels 12m and comes to a complete stop. the ball is subject to an acceleration of -1.2m/s squared. calculate original velocity, time to stop, and average velocity

Solution:

Let:

$$S = 12 \, \text{m}$$ $$a = -1.2 \, \text{m/s}^2$$

v-?, t-?, v_average -?

Such as:

$$S = \frac{1}{2} a t^2, \quad t = \sqrt{\frac{2S}{a}}$$

*acceleration by value, without " - "

$$v = a t, \quad v = \sqrt{2 a S}$$ $$v_{average} = \frac{v}{t} = \frac{\sqrt{2 a S}}{\sqrt{\frac{2S}{a}}}, \quad v_{average} = a$$

* by value

$$t = \sqrt{\frac{2 \cdot 12}{1.2}} = 4.47 \, \text{s}$$ $$v = \sqrt{2 \cdot 1.2 \cdot 12} = 5.37 \, \text{m/s}$$ $$v_{average} = 1.2 \, \text{m/s}$$