# Josh rolled a bowling ball down a lane in 2.5 s. The ball traveled at a constant acceleration of 1.8 m/s2 and was traveling at a speed of 7.6 m/s by the time it reached the pins at the end of the lane. How fast was the ball going when it left?

Feb 6, 2016

${\text{3.1 m s}}^{- 1}$

#### Explanation:

The problem wants you to determine the speed with which Josh rolled the ball down the alley, i.e. the initial speed of the ball, ${v}_{0}$.

So, you know that the ball had an initial speed ${v}_{0}$ and a final speed, let's say ${v}_{f}$, equal to ${\text{7.6 m s}}^{- 2}$.

Moreover, you know that the ball had a uniform acceleration of ${\text{1.8 m s}}^{- 2}$.

Now, what does a uniform acceleration tell you?

Well, it tells you that the speed of the object changes at a uniform rate. Simply put, the speed of the ball will Increase by the same amount every second.

Acceleration is measured in meters per second squared, ${\text{m s}}^{- 2}$, but you can think about this as being meters per second per second, ${\text{m s"^(-1) "s}}^{- 1}$. In your case, an acceleration of ${\text{1.8 m s"^(-1) "s}}^{- 1}$ means that with every second that passes, the speed of the ball increases by ${\text{1.8 m s}}^{- 1}$.

Since you know that the ball traveled for $\text{2.5 s}$, you can say that its speed increased by

2.5 color(red)(cancel(color(black)("s"))) * "1.8 m s"^(-1) color(red)(cancel(color(black)("s"^(-1)))) = "4.5 m s"^(-1)

Since its final speed is ${\text{7.6 m s}}^{- 1}$, it follows that its initial speed was

${v}_{0} = {v}_{f} - {\text{4.5 m s}}^{- 1}$

v_0 = "7.6 m s"^(-1) - "4.5 m s"^(-1) = color(green)("3.1 m s"^(-1))

You actually have a very useful equation that describes what I just did here

$\textcolor{b l u e}{{v}_{f} = {v}_{0} + a \cdot t} \text{ }$, where

${v}_{f}$ - the final velocity of the object
${v}_{0}$ - its initial velocity
$a$ - its acceleration
$t$ - the time of movement

You can double-check the result by using this equation

"7.6 m s"^(-1) = v_0 + "1.8 m s"^(-1) color(red)(cancel(color(black)("s"^(-1)))) * 2.5color(red)(cancel(color(black)("s")))

Once again, you will have

v_0 = "7.6 m s"^(-1) - "4.5 m s"^(-1) = color(green)("3.1 m s"^(-1))