Question

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?

Answer 1

`"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 `"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))`