Question

Question :-



A particle initially at rest moves along x-axis from origin. Its acceleration varies with time as a=(6t+5)m/s. The distance covered in first 2 sec is (in multiples of 6m)​

Answer 1

ans is 18 and it is 6×3.....

Answer 2

Given us the acceleration at any instant as,

• a = 6t + 5 m/s²

We have to find the distance covered in the first 2 seconds.

Also, a = dv / dt, So,

⇒ dv / dt = 6t + 5

⇒ dv = 6t + 5 dt

Integrate both sides,

⇒ ∫ dv = ∫ 6t + 5 dt

⇒ v - u = 6t² / 2 + 5t

Because the particle was initial at rest, so u = 0

⇒ v = 3t² + 5t

Now, we know v = dx / dt

⇒ dx / dt = 3t² + 5t

⇒ dx = 3t² + 5t dt

Integrate both sides, to get the relation into position w.r.t time.

⇒ ∫ dx = ∫ 3t² + 5t dt

⇒ x₂ - x₁ = 3t³ / 3 + 5t² / 2

⇒ x₂ - x₁ = t³ + 5t² / 2

We have to find the distance covered, from t = 0 to t = 2

So, Distance covered at t = 2

⇒ Distance travelled = (2)³ + 5(2)² / 2

⇒ Distance travelled = 8 + 20/2

⇒ Distance travelled = 18 m

Hence, The distance covered by the particle in first 2 seconds is 18 m.