Deceleration is a very commonly occurring phenomenon in our daily lives. Anytime we are travelling, we may notice in the vehicle that when we feel moving forward relative to the vehicle, then we are experiencing Deceleration (a slow down of our Velocity). Here, Deceleration is to be considered as a special case of Acceleration whereby it only applies to objects slowing down. In short, it is the rate at which an object slows down.
Acceleration is a Vector attribute of an object in Motion. This is because of two components : a magnitude and a direction. To denote the direction in cases of one-dimensional Motion, negative and positive signs are used. Thus, if the signs are negative then the object is decelerating or is said to be in retardation.
Deceleration can be understood as the opposite phenomenon of Acceleration. The Deceleration can be calculated by dividing the final Velocity minus the initial Velocity, by the amount of time taken for this drop in Velocity. The formula for Acceleration must be used here, with a negative sign, to identify the Deceleration value.
Deceleration, also known as retardation or negative Acceleration, is the Acceleration acts in the opposite direction of Motion and is responsible for reducing the Velocity of a body.
Deceleration(a) =
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If, u = initial Velocity
v = final Velocity
a = constant Deceleration
t = time taken
and
s = displacement
Then equations of Motion is given by:
v = u – at
s = ut – ½ at2
v2 = u2 – 2as
SI unit of Deceleration is m/s².
Example:
An object moving with a Velocity of 40m/s is brought to rest in 8 seconds by a constant Deceleration. Find the Deceleration applied.
Ans:
Here, initial Velocity(u) = 40m/s
Final Velocity(v) = 0 m/s
Deceleration = a
Time = 8 s
From 1st equation of Motion:
v = u – at
0 = 40 – a.8
a = 40/8
= 5 m/s².
Question:
If an object is accelerated from rest with an Acceleration of 2 m/s² for 10s, How much Deceleration is needed to bring the object to rest in 4s.
Deceleration in Gravity Units (G’s)
To calculate Deceleration rate when an object is under the influence of gravity, and if one desires to obtain the result in terms of gravity units (G), one of the two methods described here can be used.
First Method:
Divide the Deceleration by the standard gravitational Acceleration (the value of which is 9.8m/s²). The result gives the average number of G’s applied to the moving object to achieve the Deceleration.
We can understand this with the help of an example: Find the G Force required to stop the car with Deceleration equal to -27.66 feet per second square
The calculated Deceleration for the moving car is 27.66 feet per second per second. The Deceleration is equivalent to:
27.66/32 = 0.86 G’s
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