# Acceleration

## Practice

### practice problem 1

- A car is said to go "zero to sixty in six point six seconds". What is its acceleration in m/s
^{2}? - The driver can't release his foot from the gas pedal (a.k.a. the accelerator). How many additional seconds would it take for the driver to reach 80 mph assuming the aceleration remains constant?
- A car moving at 80 mph has a speed of 35.8 m/s. What acceleration would it have if it took 5.0 s to come to a complete stop?

#### solution

Well first of all, we shouldn't be dealing with English units. They're difficult to work with, so let's convert them straight away and then do the old "plug and chug".

*v*=60 mile 1,609 m 1 hour 1 hour 1 mile 3,600 s *v*= 26.8 m/s*a*=∆ *v*= *v*−*v*_{0}∆ *t*∆ *t**a*=26.8 m/s − 0 m/s 6.6 s *a*= 4.06 m/s^{2}Since the question asked for acceleration and acceleration is a vector quantity this answer is not complete. A proper answer must include a direction as well. This is quite easy to do. Since the car is starting from rest and moving forward, its acceleration must also be forward. The ultimate, complete answer to this problem is the car is accelerating at…

**a**= 4.06 m/s^{2}forwardWe should convert the final speed to SI units.

*v*=80 mile 1,609 m 1 hour 1 hour 1 mile 3,600 s *v*= 35.8 m/sUse 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*−*v*_{0}∆ *t*∆ *t*∆ *t*=*v*−*v*_{0}*a*∆ *t*=35.8 m/s − 26.8 m/s 4.06 m/s ^{2}∆ *t*= 2.22 sAlternate 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…∆ *v*_{2}= ∆ *t*_{2}∆ *v*_{1}∆ *t*_{1}80 mph = ∆ *t*_{2}60 mph 6.6 s ∆ *t*_{2}= 8.8 sThis is not the answer. It is the time elapsed from the moment when the car began to move. The question was about the additional time needed, so we should subtract the time required to go from zero to sixty. Thus…

∆

*t*= 8.8 s − 6.6 s = 2.2 sThe two methods give essentially the same answer.

Quite simple. Let's do it.

*a*=∆ *v*= *v*−*v*_{0}∆ *t*∆ *t**a*=0 m/s − 35.8 m/s 5.0 s *a*= −7.16 m/s^{2}Nothing surprising there except the negative sign. When a vector quantity is negative what does it mean? There are several interpretations of this, but I think mine is the best. When a vector has a negative value, it means that it points in a direction opposite that of the positive vectors. In this problem, since the positive vectors are assumed to point forward (What other direction would a normal car drive?) the acceleration must be backward. Thus the complete answer to this problem is that the car's acceleration is…

**a**= 7.16 m/s^{2}backwardAlthough it is common to assign deceleration a negative value, negative acceleration does not automatically imply deceleration. When dealing with vector quantities, any direction can be assumed positive…

up, down, right, left,

forward, backward, north, south, east, west and the corresponding opposite direction assumed negative…

down, up, left, right,

backward, forward, south, north, west, east. It won't matter which you chose as long as you are consistent throughout a problem. Don't learn any rules for assigning signs to particular directions and don't let anyone tell you that a certain direction

*must be*positive or*must*be negative.

### practice problem 2

^{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.

v_{0} = |
−40 m/s |

v = |
+50 m/s |

∆t = |
^{1}_{30} s |

a = |
? |

a = |
v−v_{0} |
= | (+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 |
||||

### practice problem 3

#### solution

Answer it.

### practice problem 4

#### solution

Answer it.