Question

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

Answer:

Explanation:

Usi a = v/t

We get area under a-t as velocity

Then use 1 st equation of motion

See from pic

Answer 2

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

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

To Find: the velocity at t = 8 s

Solution:

We know that acceleration (a) can be written as,

a = dv/dt         [ where v = velocity and t = time]

For a larger change in velocity and time, the same equation can be written as,

a = Δv / Δt

⇒ Δv = a × Δt    [ which is the area under the triangle]

now, initial velocity (u) = 10 m/s, acceleration = 2 m/s^2,

thus we first the velocity at t = 6 s,

   Δv = 1/2 ×a × Δt

⇒ v - u = 1/2 × 2 × (6 - 0)

⇒ v - 10 = 6

⇒ v = 16 m/s

∴ velocity of particle at t = 6 s is 16 m/s

Now for the rest of the journey of t = 8 s, there is deceleration ( negative sign)

again, initial velocity = 16 m/s, acceleration = -6 m/s^2,

Δv = 1/2 ×a × Δt

⇒ v - u = 1/2 × (-6) × (8 - 6)

⇒ v - 16 = - 6

⇒ v = 10 m/s

Thus the velocity of a particle at t = 8 s is 10 m/s.

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