Question

A particle moves with acceleration given as a =
8t+6 m/s2. Find displacement of particles at
t=3 sec. if at t=0 it starts from origin with initial
speed 1 m/s :-​

Answer 1

Answer:

$\sqrt{107}$

Explanation:

Answer 2

The displacement of the particle at t = 3 sec is 63 m.

Given: The relation: a = 8t + 6 m/s² and the particle starts at t = 0 from origin with initial speed 1 m/s.

To Find: The displacement of particle at t = 3 sec.

Solution:

We know that acceleration and displacement can be represented as;

              a = dv / dt                                                                          ...(1)

              v = ds / dt                                                                          ...(2)

Where a = acceleration, v = velocity, s = displacement.

Coming to the numerical, we are given;

          a = 8t + 6

From (1), we can say that;

      ⇒ a = dv / dt = 8t + 6

      ⇒  dv = ( 8t + 6 ) dt

Integrating both the sides, we shall get the velocity;

      ⇒  $\int\limits {dv} \, = \int\limits ( 8t + 6 )dt$

      ⇒  v = 8 × ( t² / 2 ) + 6t

      ⇒  v = 4t² + 6t

Now, we can say that;

      ⇒  v = ds / dt = 4t² + 6t

      ⇒  ds = ( 4t² + 6t ) dt

Integrating both the sides we get;

      ⇒ $\int\limits \, ds = \int\limits ( 4t^2 + 6t ) dt$

      ⇒ s = 4 × ( t³ / 3 ) + 6 × ( t² / 2 )

Putting t = 3 sec, in the above equation, we get;

      ⇒  s  =   4 × ( 3³ / 3 ) + 6 × ( 3² / 2 )

      ⇒  s  =  ( 4 × 9 ) + 27

      ⇒  s  = 63 m

Hence, the displacement of the particle at t = 3 sec is 63 m.

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