Question

9.
The acceleration (a)-time (t) graph of a particle
moving in a straight line ​is as shown in figure. At
time t = 0, the velocity of particle is 10 m/s. What
is the velocity at t = 8 s? [NCERT Pg. 45]
​

Answer 1

Explanation:

Case 1: Area of the triangle above the positive x axis.

Case 2: Area of the triangle below the positive x acis

Answer 2

The velocity at t = 8 s is 10 m/s.

Given: The acceleration (a)-time (t) graph of a particle.

At time t = 0, the velocity of particle is 10 m/s.

To Find: The velocity at t = 8 s.

Solution:

• We know that the acceleration of  particle can be represented as,

         a = dv/dt  

     ⇒ dv = a × dt                                

• The area of an acceleration (a)-time (t) graph gives us the velocity of the particle.

We need to find the velocity at t = 8 s when t = 0 is 10 m/s. So, initial velocity (u) = 10 m/s

The nature of the graph changes at t = 6 s, so it has to be calculated separately. We shall first find the velocity at t = 6,

        Δ V = 1/2 × 2 × ( 6 - 0 )

  ⇒ v - u = 6

  ⇒ v = (6 + 10) m/s             [ Given: u = 10 m/s ]

         = 16 m/s

Now, for the part t = 6 s and t = 8 s, initial velocity (u) = 16 m/s, so applying same concept, we get;

         Δ V = 1/2 × ( - 6 ) × ( 8 - 6 )

    ⇒ v - u = - 6

    ⇒ v = ( - 6 + 16 ) m/s

           = 10 m/s

Hence, the velocity at t = 8 s is 10 m/s.

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