Acceleration

Use the fact that change equals rate times time, and then add that change to our velocity at the end of the previous problem. Algebra will do the rest for us.

a =	∆v	=	v − v₀
	∆t		∆t

∆t =	v − v₀
	a

∆t =	35.8 m/s − 26.8 m/s
	4.06 m/s 2

∆t = 2.22 s

Alternate solution. We don't need no stinkin' conversions with this method. The ratio of eighty to sixty is a simple one, namely 4 ₃ . From our definition of acceleration, it should be apparent that time is directly proportional to change in velocity when acceleration is constant. Thus…

∆v₂	=	∆t₂
∆v₁	=	∆t₁

80 mph	=	∆t₂
60 mph	=	6.6 s

∆t₂ = 8.8 s

a =	∆v	=	v − v₀
	∆t		∆t

a =	0 m/s − 35.8 m/s
	5.0 s

a = −7.16 m/s 2

practice problem 2

A baseball is pitched at 40 m/s (90 mph) in a Major League game. The batter hits the ball on a line drive straight toward the pitcher at 50 m/s (112 mph). Determine the the acceleration of the ball if it was in contact with the bat for 1 ₃₀ s.

solution

Acceleration is the rate of change of velocity with time. Since velocity is a vector, this definition means acceleration is also a vector. When it comes to vectors, direction matters as much as size. In a simple one-dimensional problem like this one, directions are indicated by algebraic sign. Every quantity that points away from the batter will be positive. Every quantity that points toward him will be negative. Thus, the ball comes in at −40 m/s and goes out at +50 m/s. If we didn't pay attention to this detail, we wouldn't get the right answer.

v₀ =	−40 m/s
v =	+50 m/s
∆t =	1 ₃₀ s
a =	?

a =	v−v₀	=	(+50 m/s) − (−40 m/s)
	∆t		1 ₃₀ s
a =	(+90 m/s)(30 s −1 ) = +2700 m/s 2
				a = 2700 m/s 2 away from the batter