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
Feb 6, 2016

Answer:

#"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))#