Maximum velocity attained by the particle

So, the question here is -We need to find the maximum velocity achieved by the particle according to the acceleration-time graph. This question was asked on one of the IIT exams. The graph is given below: 

This is a very interesting and conceptual Kinematics problem.

As we know, velocity of a particle is given by v(t) = u+at where u = initial velocity and a = acceleration during time t. The assumption in this equation is that a = constant. However, in above problem,we see that acceleration is not constant. It’s varying with time. But interesting thing to note here is that the rate of change of acceleration is constant. So, we have the following equation:

da/dt = slope of graph.

From the graph we can see that in this case, the slope is -10/11(-ve sign as the line is going downwards)

Therefore, we had the following equation:
da/dt = -10/11-----------------------(I)

Now all we need is to integrate it which gives us
∫da = ∫(-10)/(11)dt or a =  (-10)/(11)  +c  where c is a constant which we need to find.

At t = 0, a=10 ∴  c = 10
We had our first equation -

a = (-10)/(11)t + 10. Since velocity is also continuously changing, we can write it as 

dv/dt = (-10)/(11)t +10 ………………………………….(II)

Again integrating it we get,

v = -(10/22) t²  + 10t +m where m = constant of integration

Again, for finding the value of m, at t = 0, v =0 since the particle starts from rest.

It gives m =0 ∴ v= -(10/22)t² +10t ………………….(III)

So now we had got the velocity as a function of time. Only thing we need to know now is to find the maximum velocity achieved by the particle.

Now, we know that the velocity can be maximum or minimum when its derivative becomes 0. From equation (II) above, we get

dv/(dt) = 0 i.e.  (-10)/(11)t + 10 = 0 => t = 11 seconds, so substituting this value in equation (III), we get

v =(-10)/(22).(11).(11) + 10.11 = 55m/s 

So this velocity can be maximum or minimum. Since the particle starts from rest this can’t be minimum, hence maximum velocity attained = 55m/s

Now though we had found the solution lets analyze the problem a bit more. The particle starts from rest and attains the maximum velocity at t =11s and is 55 m/s. Let’s analyze its motion after it has attained maximum velocity and assume that the da/dt remains constant as in figure. Let’s find when the velocity will be 0 again. From equation (III),

 -(10/22)t² + 10t = 0 => t=0, t=0, 22 seconds i.e. velocity becomes 0 again at t = 22 seconds. After that it will keep on increasing but in opposite direction. Let’s see the graph of the velocity versus time:

Plotting a graph gives us a good idea about the motion of the particle and it is evident from the values we got from solving the equations. Again writing this equation in terms of position of particle we get, from equation (III),

dx/dt= -(10/22)t² + 10t => x = -(10/66)t³+5t²

We can find out the distance travelled by particle from the above equation. So, from the instant the particle starts and again comes to rest, it has travelled  x= -(10/66) (22)³ +5(22)² = Around 800 meters.Below is the graph of the particle’s position with respect to time:

So we see how a simple problem can be analyzed conceptually and understood in terms of pure fundamentals. In next few sessions, we will delve into some more harder kinematics problems.