Objective: Know that a deceleration is a negative acceleration and use this in calculations
In physics, acceleration (\$a\$) is the rate of change of velocity with time:
\$a = \frac{\Delta v}{\Delta t}\$
When the velocity of an object decreases, the acceleration is directed opposite to the motion. This is called deceleration and is simply a negative acceleration:
\$a_{\text{decel}} = -\,|a|\$
Key Points
Deceleration is not a different physical quantity; it is just acceleration with a negative sign.
The magnitude of deceleration is the absolute value of the acceleration.
In equations, always keep the sign consistent with the chosen direction (positive forward, negative backward).
Typical situations: braking a car, a ball rolling up a slope, an object thrown upwards.
Symbols and Units
Symbol
Quantity
Unit
\$v\$
velocity
m s\(^{-1}\)
\$\Delta v\$
change in velocity
m s\(^{-1}\)
\$t\$
time
s
\$\Delta t\$
change in time
s
\$a\$
acceleration (positive) or deceleration (negative)
m s\(^{-2}\)
Worked Example
A car travelling at \$20\ \text{m s}^{-1}\$ brakes uniformly to a stop in \$5\ \text{s}\$. Find the acceleration and state whether it is a deceleration.
Identify the initial and final velocities:
\$v_i = 20\ \text{m s}^{-1}\$ (forward)
\$v_f = 0\ \text{m s}^{-1}\$
Calculate the change in velocity:
\$\Delta v = vf - vi = 0 - 20 = -20\ \text{m s}^{-1}\$
Interpretation: The acceleration is \$-4\ \text{m s}^{-2}\$, i.e. a deceleration of \$4\ \text{m s}^{-2}\$ opposite to the direction of motion.
Another Example – Object Thrown Upwards
A ball is thrown vertically upward with an initial speed of \$15\ \text{m s}^{-1}\$. Taking upward as positive, the acceleration due to gravity is \$-9.8\ \text{m s}^{-2}\$. Determine the time taken to reach the highest point.
\$v = u + at\$
At the highest point \$v = 0\$, so:
\$0 = 15 + (-9.8)t\$
\$t = \frac{15}{9.8} \approx 1.53\ \text{s}\$
The negative sign of \$a\$ indicates a deceleration of the upward motion.
Practice Questions
A cyclist traveling at \$8\ \text{m s}^{-1}\$ slows down uniformly to \$2\ \text{m s}^{-1}\$ in \$4\ \text{s}\$. Calculate the acceleration and state whether it is a deceleration.
A train decelerates at \$1.2\ \text{m s}^{-2}\$ for \$30\ \text{s}\$. If its initial speed is \$25\ \text{m s}^{-1}\$, what is its final speed?
A ball rolls up a gentle slope, slowing from \$3\ \text{m s}^{-1}\$ to rest in \$6\ \text{s}\$. Determine the magnitude of the deceleration.
Explain why the sign of acceleration must be considered when using the equation \$s = ut + \frac{1}{2}at^{2}\$ for a decelerating object.
Common Mistakes to Avoid
Treating deceleration as a separate quantity and forgetting the negative sign.
Using the magnitude of acceleration in equations that require a signed value.
Mixing up the chosen positive direction; always keep it consistent throughout a problem.
Suggested diagram: A car braking to a stop, showing initial velocity, final velocity, and the negative acceleration vector.
Summary
Deceleration is simply acceleration with a negative sign, indicating that the velocity is decreasing in the chosen positive direction. By keeping track of signs, you can correctly apply the kinematic equations to any situation involving slowing down.