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?

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 `"7.6 m s"^(-2)`.

Moreover, you know that the ball had a uniform acceleration of `"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, `"m s"^(-2)`, but you can think about this as being meters per second per second, `"m s"^(-1) "s"^(-1)`. In your case, an acceleration of `"1.8 m s"^(-1) "s"^(-1)` means that with every second that passes, the speed of the ball increases by `"1.8 m s"^(-1)`.

Since you know that the ball traveled for `"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 `"7.6 m s"^(-1)`, it follows that its initial speed was

`v_0 = v_f - "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

`color(blue)(v_f = v_0 + a * t)" "`, 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))`