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
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
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,
Since you know that the ball traveled for
#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))#