**In mechanics we are interested in trying to understand the motion of objects. In this chapter, the motion of objects in 1 dimension will be discussed. Motion in 1 dimension is motion along a straight line.**

## 2.1. Position

**The position of an object along a straight line can be uniquely identified by its distance from a (user chosen) origin. (see Figure 2.1).**

**Note**: the position is fully specified by 1 coordinate (that is why this a 1 dimensional problem).

Figure 2.1. One-dimensional position.

Figure 2.2. x vs. t graphs for various velocities.

For a given problem, the origin can be chosen at whatever point is
convenient. For example, the position of the object at time t = 0 is often
chosen as the origin. The position of the object will in general be a function
of time: x(t). Figure 2.2. shows the position as a function of time for an
object at rest, and for objects moving to the left and to the right.The slope of the curve in the position versus time graph depends on the velocity of the object. See for example Figure 2.3. After 10 seconds, the cheetah has covered a distance of 310 meter, the human 100 meter, and the pig 50 meter. Obviously, the cheetah has the highest velocity. A similar conclusion is obtained when we consider the time required to cover a fixed distance. The cheetah covers 300 meter in 10 s, the human in 30 s, and the pig requires 60 s. It is clear that a steeper slope of the curve in the x vs. t graph corresponds to a higher velocity.

Figure 2.3. x vs. t graphs for various creatures.

## 2.2. Velocity

**An object that changes its position has a non-zero velocity. The**

**average velocity**of an object during a specified time interval is defined as:

_{ }

_{1}and at time t = t

_{2}. This is nicely illustrated in

**sample problem 2-1 and 2-2**.

**Sample Problem 2-1**

You drive a beat-up pickup truck down a straight road for 5.2 mi at 43 mi/h, at which point you run out of fuel. You walk 1.2 mi farther, to the nearest gas station, in 27 min (= 0.450 h). What is your average velocity from the time you started your truck to the time that you arrived at the station ?

The pickup truck initially covers a distance of 5.2 miles with a velocity of 43 miles/hour. This takes 7.3 minutes. After the pickup truck runs out of gas, it takes you 27 minutes to walk to the nearest gas station which is 1.2 miles down the road. When you arrive at the gas station, you have covered (5.2 + 1.2) = 6.4 miles, during a period of (7.3 + 27) = 34.3 minutes. Your average velocity up to this point is:

_{ }

**Sample Problem 2-2**

Suppose you next carry the fuel back to the truck, making the round-trip in 35 min. What is your average velocity for the full journey, from the start of your driving to you arrival back at the truck with the fuel ?

It takes you another 35 minutes to walk back to your car. When you reach your truck, you are again 5.2 miles from the origin, and have been traveling for (34.4 + 35) = 69.4 minutes. At that point your average velocity is:

_{ }

_{ }

**instantaneous velocity**of an object at a given instant. The instantaneous velocity is the value that the average velocity approaches as the time interval over which it is measured approaches zero:

_{ }

**sample problem 2-5**.

_{ }

_{ }

_{ }

## 2.3. Acceleration

**The velocity of an object is defined in terms of the change of position of that object over time. A quantity used to describe the change of the velocity of an object over time is the acceleration a. The**

**average acceleration**over a time interval between t

_{1}and t

_{2}is defined as:

_{ }

**instantaneous acceleration**a is defined as:

_{ }

_{ }

_{ }

## 2.4. Constant Acceleration

**Objects falling under the influence of gravity are one example of objects moving with constant acceleration. A constant acceleration means that the acceleration does not depend on time:**

_{ }

_{ }

_{0}is the velocity of the object at time t = 0. From the velocity, the position of the object as function of time can be calculated:

_{ }

_{0}is the position of the object at time t = 0.

**Note 1**: verify these relations by integrating the formulas for the position and the velocity.

**Note 2**: the equations of motion are the basis for most problems (see sample problem 7).

**Sample Problem 2-8**

Spotting a police car, you brake a Porsche from 75 km/h to 45 km/h over a distance of 88m. a) What is the acceleration, assumed to be constant ? b) What is the elapsed time ? c) If you continue to slow down with the acceleration calculated in (a) above, how much time would elapse in bringing the car to rest from 75 km/h ? d) In (c) above, what distance would be covered ? e) Suppose that, on a second trial with the acceleration calculated in (a) above and a different initial velocity, you bring your car to rest after traversing 200 m. What was the total braking time ?

a) Our starting points are the equations of motion:

_{ }

(1)

_{ }

(2)

The following information is provided:* v(t = 0) = v

_{0}= 75 km/h = 20.8 m/s

* v(t

_{1}) = 45 km/h = 12.5 m/s

* x(t = 0) = x

_{0}= 0 m (Note: origin defined as position of Porsche at t = 0 s)

* x(t

_{1}) = 88 m

* a = constant

From eq.(1) we obtain:

_{ }

(3)

Substitute (3) in (2):_{ }

(4)

From eq.(4) we can obtain the acceleration a:_{ }

(5)

b) Substitute eq.(5) into eq.(3):_{ }

(6)

c) The car is at rest at time t_{2}:

_{ }

(7)

Substituting the acceleration calculated using eq.(5) into eq.(3):_{ }

(8)

d) Substitute t_{2}(from eq.(8)) and a (from eq.(5)) into eq.(2):

_{ }

(9)

e) The following information is provided:* v(t

_{3}) = 0 m/s (Note: Porsche at rest at t = t

_{3})

* x(t = 0) = x

_{0}= 0 m (Note: origin defined as position of Porsche at t = 0)

* x(t

_{3}) = 200 m

* a = constant = - 1.6 m/s

^{2}

Eq.(1) tells us:

_{ }

(10)

Substitute eq.(10) into eq.(2):_{ }

(11)

The time t_{3}can now easily be calculated:

_{ }

(12)

## 2.5. Gravitational Acceleration

A special case of constant acceleration is free fall (falling in vacuum). In problems of free fall, the direction of free fall is defined along the y-axis, and the positive position along the y-axis corresponds to upward motion. The acceleration due to gravity (g) equals 9.8 m/s

^{2}(along the negative y-axis). The equations of motion for free fall are very similar to those discussed previously for constant acceleration:

_{ }

_{ }

_{ }

_{0}and v

_{0}are the position and the velocity of the object at time t = 0.

**Example**

A pitcher tosses a baseball straight up, with an initial speed of 25 m/s. (a) How long does it take to reach its highest point ? (b) How high does the ball rise above its release point ? (c) How long will it take for the ball to reach a point 25 m above its release point.

Figure 2.4. Vertical position of baseball as function of time.

a) Our starting points are the equations of motion:_{ }

_{ }

* v(t = 0) = v

_{0}= 25 m/s (upwards movement)

* y(t = 0) = y

_{0}= 0 m (Note: origin defined as position of ball at t = 0)

* g = 9.8 m/s

^{2}

The highest point is obtained at time t = t

_{1}. At that point, the velocity is zero:

_{ }

_{ }

b) The position of the ball at t

_{1}= 2.6 s can be easily calculated:

_{ }

_{ }

_{ }

t = 1.4 s

t = 3.7 s

Figure 2.5. Velocity of the baseball as function of time.

The velocities of the ball at these times are (see also Figure 2.5):
v(t = 1.4 s) = + 11.3 m/s

v(t = 3.7 s) = - 11.3 m/s

At t = 1.4 s, the ball is at y = 25 m with positive velocity (upwards
motion). At t = 2.6 s, the ball reaches its highest point (v = 0). After t =
2.6 s, the ball starts falling down (negative velocity). At t= 3.7 s the ball
is located again at y = 25 m, but now moves downwards.
## 0 comments:

## Post a Comment