Deceleration is acceleration in reverse; whereas acceleration is the rate at which an object speeds up, deceleration is the rate at which it slows down. For example, an airplane screeching to a halt must have a high deceleration rate in order 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 involves the time it takes to slow the object and the other, the distance. Calculated rates of deceleration can be expressed in units of standard earth gravity (G’s).

## 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 you will calculate. 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 it to feet per second by multiplying the result by 1.47 (5,280 feet per mile/3,600 seconds per hour). Similarly, multiply kilometers per hour by 0.278 to convert it to meters per second.

Divide the speed change by the time during which the change occurred. This is 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.

The speeds are 300 x 1.47 = 440 feet per second, and 60 x 1.47 = 88 feet per second. The speed reduction is 300 – 88 = 212 feet per second. The rate of deceleration is 212/30 = 7.07 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.

60 mph is 88 feet per second. Because the ending speed is zero, the difference is this result squared: 7,744 feet squared per second squared. The deceleration rate is 7,744/(2 x 140) = 27.66 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 it is 9.8 meters per second per second. The result is the average number of G’s applied to achieve the deceleration.

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

The deceleration was calculated to be 27.66 feet per second per second. This is equivalent to 27.66 / 32 = 0.86 G’s.

#### TL;DR (Too Long; Didn't Read)

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.