Question

The velocity of a particle moving along a straight line as a function of time is v = 2t² – 2t, where vis in m/s andy is in second. The time interval for which the particle continuously decelerates is
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Answer 1

Answer:

We know that acceleration is given by differentiating velocity with respect to time. Hence differentiating the given function we get:

⇒ v = 2t² - 2t

⇒ a = dv/dt

⇒ a = d ( 2t² - 2t ) / dt = 4t - 2  

Hence acceleration is given as: 4t - 2.  → Eqn. 1

Differentiating once again, we get the extremum of the function.

⇒ da/dt = 4

Since 4 is greater than zero, the value of a obtained on taking the value of 't' to be zero would give us the minimum value.

Substituting t = 0, we get:

⇒ a = 4(0) - 2 = 0 - 2 = -2

Hence a = -2 m/s²

Now equating a = 0, we get 't' to be:

⇒ 0 = 4t - 2

⇒ 2 = 4t

⇒ t = 2/4 = 0.5 seconds

Hence at t = 0.5 seconds, acceleration becomes zero after which it starts to increase.

Hence between the time interval t = 0 sec to t = 0.5 sec, we see that the particle undergoes continuous deceleration.