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# Motion in One Dimension Study Guide (page 2)

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Updated on Sep 26, 2011

#### Example 2

Suppose a car starts from rest. The car accelerates straight down the road and at t = 30 s attains a speed of v = 30 km/h. Determine the average acceleration of the car.

#### Solution 2

(30 km/h – 0 km/h)/(30 s – 0 s) = 1.0(km/h)/s = 1,000 m/3,600 S2 0.28 m/s2

Similar to the case of velocity, we start by defining an average acceleration over a finite time interval Δt and then get to the concept of instantaneous acceleration, taking this time interval to zero:

Whenever the acceleration and velocity vectors have opposite directions, the object slows down and is said to be decelerating.

## Graphs and Motion

An easy and clear method of quickly analyzing relationships among large quantities of numerical data is to plot the experimental data as a function of time or of some other variable. Straight lines on a graph can be easily recognized and interpreted, and they play an important role in graphically analyzing experimental data.

By far, the most usual kind of graph is the Cartesian one, where both axes have linear scales. The horizontal axis is also called abscissa, and the vertical axis is called ordinate. The graph shows the dependency of the quantity, which is represented as the ordinate on the qualitity, which is represented as the abscissa.

We define the slope of a straight line on the graph by choosing any two points on the line and taking the ratio of the ordinate separation of these two points to their abscissas separation. If the graph is not made up of straight lines, then we can define a slope at any point along the curve by drawing a tangent to the curve and calculating the slope of this tangent, using the slope definition for a straight line.

In the case of position versus time graphs, the slope of the graph at a certain time is immediately associated with the instantaneous velocity at that time, as the ratio of position coordinate difference versus the time interval to go from the initial position to the other is an exact reformulation of the definition of speed. By similar reasoning, the slope of a velocity versus time graph is associated with the instantaneous acceleration of the object at that time.

#### Example

The two graphs in Figure 3.1 show the position versus time for two swimmers racing across a 35-meter pool. Interpret the information in the graphs.

#### Solution

There is a lot of information we can read from these graphs.

The first athlete starts at time t = 0 and swims the first 25 meters of the race at constant speed v0 = 25 m/15 s = 1.6 m/s. He gets tired and goes the last 10 meters of the race at a much slower pace, at a speed v = 10 m/20 s = 0.5 m/s.

The second athlete starts with a handicap of 5 seconds and initially goes slowly. For example, from point C to point D, his speed is almost constant. His speed at point D, after 35 seconds from the beginning of the race, is given by the slope of the tangent and is vD = 7.5 m/2s s = 0.3 m/s. But at point E, 50 seconds after the beginning of the race, his instantaneous speed increased to vE = 16.5 m/15 s = 1.1 m/s. The last 10 seconds of the race, he rides Flipper the dolphin, and at point F, his final speed is around 2.8 m/s. This reminds us somewhat of the Hare and Tortoise fable but with an aquatic spin!

## Motion with Constant Acceleration

Let's examine now the case of constant acceleration motion. This is an important kind of motion encountered in frequent approximations to real situations. An object sliding down an incline or multiple objects falling near the surface of the earth when we neglect air resistance are both situations where we encounter constant acceleration motion.

As the acceleration is constant, the instantaneous and average accelerations are equal, and we can invert the definition to get:

Δv = a Δt

If we consider that at time t = 0, the initial velocity is v0, we can write this as:

v (t) = v0 + a · t

For the case of the one-dimensional motion, this becomes a scalar relation between speeds, with a positive acceleration a when it is directed in the sense of the motion, and a negative acceleration a when its direction is against the direction of the motion.

If we consider that at the initial time t = 0, there is an initial displacement x0, then the position at time t is given by:

Solving for t in the velocity equation and inserting the result in the displacement equation, we obtain another useful relation that is independent of time:

v2 = v20 + 2aΔx

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