When we talk about constant entities, there must be a steady change in the terms involved in the process. We know how to plot a constant acceleration graph from the previous post, but how to find constant acceleration with velocity and time?

**To find constant acceleration with velocity and time, the motion of the particle has to be linear. Uniform increase or decrease in the velocity over time provides useful stuff for finding the constant acceleration. This post is mainly concerned with how to find constant acceleration with velocity and time in different aspects and approaches.**

To calculate constant acceleration, we need to assume certain things like a change in velocity have to be constant, and the motion should be in one dimensional. These assumptions help us to find the constant acceleration.

**How to Find Constant Acceleration with Velocity and Time?**

The steady change in the velocity with a regular time interval gives the constant acceleration. When a particle is in uniform linear motion, the change in the velocity with time is unvarying; then, we can find the constant acceleration.

**In order to find the constant acceleration with velocity and time, we have to take the derivative of the velocity with respect to time because a function can easily be proven to be constant by using the differential calculus method. Thus we take the velocity as a function of time to find the derivative to achieve constant acceleration quickly.**

The acceleration of the particle is given by

a = v/t

**Case (i) Since we are dealing with constant acceleration, then we must consider a steady change in velocity. **

The change in the velocity is given by

a = Δv/Δt

The change in velocity means the particle must attain more than one velocity. So let us assume that initially, the particle is moving with velocity v_{0} at time t=0, and in the next instance, the velocity has increased to v at time t. The velocity of the particle then increases v times for every successive time interval so that the acceleration will be constant throughout the motion.

Now the change in velocity can be rewritten as

a = v – v_{0}/t_{0}

From the above equation, we get the equation of motion as

v = v_{0} +at

We have taken initial velocity as zero, hence equation will be

v = 0 +at

v = at

Rearranging them for our convenience to find the constant velocity as

a = v/t

Now we have again achieved the general expression of acceleration. Let us differentiate the above equation.

a = dv/dt

The acceleration is said to be constant when we get the derivative of the above equation as a non-zero number. i.e.,

If dv/dt = constant, then a=constant.

This can also be interpreted on the graph; know more about **the graphical representation of constant acceleration with velocity and time. **As the slope of the velocity-time graph gives the acceleration.

You can understand the above expression easily by solving an example. Let us solve an example. If we consider a motion of an object in a plane and its velocity is given by v(t) = 4t-4, the acceleration is constant or not can be solved by the derivative equation. The equation of acceleration is given by

a = dv/dt

We know v(t) = 4t-4,

Differentiating the above equation with respect to t,

dv/dt =4(1)-0

dv/dt =4

Thus the time derivative of the velocity is a constant number; hence the acceleration is constant in the given velocity function.

**Case (ii) Suppose initial velocity is not zero, then the constant acceleration can be given as**

v = v_{0 }+at

Differentiating the above term

dv/dt =dv_{0}/dt + a

a=dv/dt-=dv_{0}/dt = constant

Thus difference between the first order derivative of initial and final velocity with respect to time should be not be zero and it must be a constant number.

This can be given by the graph as below

**Example Problems on How to find constant acceleration with velocity and time**

**Problem 1) An airplane is moving on the runway with an initial velocity of 76 m/s. After 28 seconds, its velocity is 82 m/s. after that, the velocity of the airplane changes consistently for every 28 seconds before it could take off. Calculate the change in velocity and constant acceleration of the airplane.**

**Solution:**

Data provided for the calculation – the initial velocity of the airplane v0 = 76 m/s.

Final velocity of the airplane v = 82 m/s.

The time taken by the airplane to attain final velocity t = 28 seconds.

The change in the velocity

∆v = v-v_{0}

∆v = 82 – 76

∆v = 6 m/s.

This implies that for every 28 seconds, the airplane increases its velocity by 6m/s.

The constant velocity is given by

a = ∆v/∆t

Substituting the value of ∆v, we get

a=6/28

a=0.21 m/s^{2}.

**Problem 2) does the acceleration is constant or not if the initial velocity of a moving object is given by v0(t)= 5t-6 and the final velocity is given as v(t) = 7t+5.**

**Solution:**

Given –initial velocity and the final velocity is given as a function of time t. Thus we need to find the acceleration as follows.

The initial velocity of the moving object is given by

v_{0}(t) = 5t-6

Now, differentiate with respect to ‘t’

dv_{0}/dt=5(1)-0

dv_{0}/dt =5

The final velocity of the object is given by

V(t) = 7t+5

Differentiate the above function with respect to t

dv/dt =7(1)+0

dv/dt =7

The constant acceleration is given by

a=dv/dt – dv_{0}/dt

Substituting the initial and final value of velocity function we get

a = 7-5

a = 2 m/s^{2}.

Thus derivative of the velocity function is constant, and the constant acceleration of the given velocity function is 2 m/s2.

**Problem 3) A jogger is jogging in a park with a certain velocity. For every 12 seconds, his speed of jogging is increased by 2 m/s; find the constant acceleration of the jogger.**

**Solution:**

Data provided for the calculation – change in velocity of the jogger v= 2 m/s.

Time is taken by the jogger to increase the speed of jogging t = 12s.

The constant acceleration is given by

a=Δv/Δt

a = 2/12

a = 0.166 m/s^{2}.

**Problem 4) How to find constant acceleration with velocity and time if the initial velocity of a moving body is given by the function of time as v0(t) = t**^{2}-6t+5

^{2}-6t+5

**Solution:**

The initial velocity is given by

v_{0}(t) = t^{2}-6t+5

Differentiating the above equation we get

dv_{0}/dt=2t-6+0

dv_{0}/dt=2t-6

From the above equation, the acceleration is not constant. There is fluctuation in the velocity with time. Hence we get the value of acceleration as a function of time.

a = 2t-6 m/s^{2}.

**Problem 5) A boat is moving with a velocity of 45 m/s at time t=23 seconds, and the same boat has a velocity of 64 m/s at time t=56 seconds. Calculate the change in velocity and time and hence find the constant acceleration.**

**Solution:**

Data given for the calculation –velocity of the boat at v_{1} = 45 m/s

Velocity of the same boat v_{2} = 64 m/s.

Time taken by the boat t_{1} to achieve velocity v_{1}= 23s.

Time taken by the same boat t_{2} to achieve velocity v_{2} = 56s.

The change in the velocity

∆v = v_{2} – v_{1}

∆v = 64 – 45

∆v = 19 m/s.

The change in the time

∆t = t_{2}-t_{1}

∆t = 56-23

∆t = 33s.

The boat increases its velocity for every 33 seconds by 19m/s.

So the acceleration of the given velocity and time is

a = Δv/Δt = 19/33

a = 0.575 m/s^{2}.