Acceleration Formula

When a stationary car starts suddenly, we get pushed up backward, and when brakes are applied, we get pushed forward against our seat, or when our car takes a sharp right turn, we get pushed towards the left. We experience these situations because our car is accelerating.

Simply when there is a change in Velocity, there will be Acceleration. Let’s understand the concept of Acceleration with illustrative examples.

Let’s suppose I have a car moving with a constant Velocity of 90 kmph along a straight line. I can see a helicopter flying at roughly a speed of 20,000 kmph. If I were to ask you that in these two cases, where do you find the Acceleration? Your answer will be surely no because both are moving at a constant pace, so no Acceleration in both cases. 

Now, if I ask you that Acceleration is equal to high speed. What will be your answer? You may say yes, but that’s not true for sure.  Want to know why?  It’s because Acceleration is the rate of change of Velocity. Now, let’s understand the Acceleration formula.

General Formula of Acceleration

We already know that Velocity is a speed with direction; therefore, it is a vector quantity. The Acceleration ‘a’ is given as:

      \[ a  = \frac{\text{Change in Velocity}}{\text{Time Taken}}\]

This formula states that the rate of change in Velocity is the Acceleration, or if the Velocity of an object changes from its initial value ‘u’ to the final value ‘v’, then the expression can be simply written as:

                         \[a =  \frac{(v - u)}{t}\]        

Acceleration Formula in Physics 

In Physics , Acceleration is described as the rate of change of Velocity of an object, irrespective of whether it speeds up or slows down. If it speeds up, Acceleration is taken as positive and if it slows down, the Acceleration is negative. It is caused by the net unbalanced force acting on the object, as per Newton’s Second Law. Acceleration is a vector quantity as it describes the time rate of change of Velocity, which is a vector quantity. Acceleration is denoted by a. Its SI unit is \[\frac{m}{s^{2}}\] and dimensions are \[[M^{0}L^{1}T^{–2}]\].

If \[v_{0}, v_{t}\] and t represents the initial Velocity, final Velocity and the time taken for the change in Velocity, then, the Acceleration is given by:

\[\overrightarrow{a} = \frac{\overrightarrow{v_t} - \overrightarrow{v_0}}{t}\]

In one dimensional motion, we can use;

\[a = \frac{v_t - v_0}{t}\]

Acceleration Formula

If \[\overrightarrow{r} \]represents displacement vector and \[\overrightarrow{v} = \frac{\overrightarrow{\text{d}r}}{\text{d}t}\] represents the velocity, then;

Acceleration:  \[\overrightarrow{a} = \frac{\overrightarrow{\text{d}v}}{\text{d}t} = \frac{\overrightarrow{\text{d}^{2}v}}{\text{d}^{2}t}\]

In one dimensional motion, where x is the displacement, and \[v = \frac{\text{d}r}{\text{d}t}\] is the Velocity, then;

\[a = \frac{\text{d}{v}}{\text{d}t} = \frac{\text{d}^{2}x}{\text{d}^{2}t}\]

Example 1:

A car starts from rest and achieves a speed of 54 \[\frac{km}{h}\] in 3 seconds. Find its Acceleration?

Solution:  \[v_0\] = 0, \[v_t\] = 54 \[\frac{km}{h}\] = 15 \[\frac{m}{s}\], t = 3s, a = ?

Acceleration:  

\[a = \frac{v_t - v_0}{t} = \frac{15 - 0}{3} = 5 \frac{m}{s^{2}}\] 

Example 2:

A body moves along the x- axis according to the relation \[x = 1 – 2 t + 3t^{2}\], where x is in meters and t is in seconds. Find the Acceleration of the body when t = 3 s

Solution:

We have: \[x = 1 – 2 t + 3t^{2}\]

then; Velocity \[v = \frac{\text{d}x}{\text{d}t} = -2 + 6t \]

Acceleration:  \[v = \frac{\text{d}v}{\text{d}t} =  6  \frac{m}{s^{2}}\].

(We see that the Acceleration is constant here. Therefore, at t = 3s also, its value is 6  \[\frac{m}{s^{2}}\]).

Solved Questions Using Acceleration Formula:

1. What will be the Acceleration of a Car if it Slows from 90 \[\frac{km}{h}\] to a Stop in 10 sec? 

Here, u = 90 \[\frac{km}{h}\] = \[ \frac{90 \times 5}{18} = 25 \frac{m}{s^{2}} \] because initially it was moving at a speed of 90 kmph then reached zero.

Final Velocity ‘v’ = 0 kmph, and t  = 10 seconds

Now, applying the formula here:

\[a  = \frac{(0 - 25)}{10}  = (-) 2.5 \frac{m}{s^{2}} \]

2. A Girl Starts her Motion in a Straight Line at a Velocity of 30 \[\frac{m}{s}\], her Velocity is Changing at a Constant Rate. If She Stops after 60 s, What is her Acceleration?

Answer: Here, the initial Velocity of a girl was 30 \[\frac{m}{s}\] and stops, so her final Velocity will become 0 m/s. Now, the deceleration or retardation occurs, which is just the opposite of Acceleration and it can be determined as:

                  \[  a  = \frac{(0 - 30)}{60} = (-) 0.5 \frac{m}{s^{2}} \]

Question 3: A Car Moves in a Circular Track with a Constant Velocity; will it Experience Acceleration?

Answer: Here, the speed is constant; however, the direction is continuously varying, which means the Velocity is also varying. It states that the car will experience Acceleration.

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