Question

A particle is moving along a straight line with variable acceleration given as, a=(4t+3)m/s^2 Where t is in seconds. The initial velocity is 10m/s. The instantaneous velocity at t = 2 seconds

Answer 1

Answer:

24 m/s

Explanation:

Here

a = 4t + 3

or dv/dt = 4t + 3

or dv = (4t + 3)dt

Integrating on both side, we get

v = 2t^2 + 3t + C

We know that v(t=0) = 10m/s

so, 10 = 2*0^2 + 3*0 + C

or C = 10

Therefore, v = 2t^2 + 3t + 10

At t = 2 sec, v = 2*2^2 + 3*2 +10

or v = 24 m/s

Answer 2

Given,

Variable acceleration of a particle=(4t+3)m/s²

Initial velocity=10m/s

To Find,

Inst. velocity at t=2 seconds.

Solution,

Here it is mentioned that, a = 4t + 3m/s²

Which means, dv/dt = 4t + 3

                        ⇒dv = (4t + 3)dt

After using integration on both side, we get,

               v = 2t² + 3t + C

We know that initial velocity given = 10m/s means t=0

    So, 10 = 2×0^2 + 3×0 + C

           ⇒C = 10

So, v = 2t² + 3t + 10

At t = 2 sec,

v = 2×2² + 3×2 +10

⇒v = 24 m/s

Hence, the Inst. velocity at t=2 seconds is 24m/s.