How to Calculate Deceleration

How to Calculate Deceleration
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Deceleration really means acceleration in reverse; whereas acceleration means the rate at which an object speeds up, deceleration means the rate at which an object slows down. For example, an airplane screeching to a halt must have a high deceleration rate to stay on the runway, and an automobile must sometimes decelerate at a precise rate to stay in the flow of traffic. Two equations are useful for calculating deceleration. One formula involves the time required to slow the object and the other formula uses the distance. Calculated rates of deceleration can be expressed in units of standard earth gravity (G’s).

Deceleration may be calculated as change in speed over a period of time by using the formula final speed (sf) minus initial speed (si) divided by the time of the change in speed (t):

\text{deceleration}=\frac{s_f-s_i}{t}

Deceleration may also be calculated as change in speed over distance by using the formula final speed squared (sf2) minus initial speed squared (si2) divided by twice the distance (d):

\text{deceleration}=\frac{s_f^2-s_i^2}{2d}

Convert units, if necessary, to be sure that the units, whether feet per second or meters per second, remain consistent.

Using Speed Difference and Time

    Subtract the ending speed from the starting speed.

    Convert the speed difference to units of speed that are compatible with the acceleration to be calculated. Acceleration is commonly expressed in feet per second per second, or meters per second per second. If the speed is in miles per hour, convert that speed to feet per second by multiplying the result by 1.47 (5,280 feet per mile divided by 3,600 seconds per hour). Similarly, multiply kilometers per hour by 0.278 to convert the speed to meters per second.

    Divide the speed change by the time during which the change occurred. This calculation yields the average deceleration rate.

    Calculate, as an example, the deceleration required to slow a landing aircraft from 300 mph to 60 mph in 30 seconds.

    Convert the speeds so:

    300 \times 1.47 = 440 \text{ feet per second, and }60 \times 1.47 = 88 \text{ feet per second}

    The speed reduction equals 300 – 88 = 212 feet per second. The rate of deceleration calculates as:

    \frac{212}{30} = 7.07 \text{ feet per second per second}

Using Speed Difference and Distance

    Convert the initial and final speeds to units that will be useful for calculating the acceleration (feet per second or meters per second). Also make sure the distance over which the speed change occurs is in a compatible unit (feet or meters).

    Square the initial speed and the final speed.

    Subtract the square of the final speed from the square of the initial speed.

    Divide by two times the distance. This is the average deceleration rate.

    Calculate, as an example, the deceleration required to stop a car in 140 feet if it is traveling 60 mph.

    Convert 60 mph to 88 feet per second. Because the ending speed equals zero, the difference is this result squared: 7,744 feet squared per second squared. The deceleration rate is:

    \frac{7744}{2\times 140}=27.66\text{ feet per second per second}

Deceleration in Gravity Units (G’s)

    Calculate the deceleration rate using one of the two methods described above.

    Divide the deceleration by the standard gravitational acceleration. In U.S. units, this is approximately 32 feet per second per second. For metric units the standard gravitational acceleration is 9.8 meters per second per second. The result gives the average number of G’s applied to achieve the deceleration.

    Enhance understanding by considering an example: Find the G force required to stop the car in the previous example.

    The calculated deceleration equaled 27.66 feet per second per second. The deceleration is equivalent to:

    \frac{27.66}{32} = 0.86\text{ G’s}

    Things You'll Need

    • Pencil and paper
    • Calculator

    Tips

    • Calculations of deceleration, like the ones in the examples, often involve only linear motion. For accelerations involving two and three dimensions, the mathematics involve vectors, which are directional, and are more complex.

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