# 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?

##### 1 Answer

#### Answer:

#### 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,

So, you know that the ball had an **initial speed** **final speed**, let's say

Moreover, you know that the ball had a **uniform acceleration** of

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*, *meters per second per second*, **every second** that passes, the speed of the ball increases by

Since you know that the ball traveled for *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

#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

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))#