Question

The velocity of the particle at any time t is given by \[v = 2t(3 - t)m{s^{ - 1}}\] . At what time is its velocity maximum?
(A) 2 s
(B) 3 s
(C) $\dfrac{2}{3}$ s
(D) \[\dfrac{3}{2}\] s

Answer 1

Hint The fastest method is to use trial and error method by substituting the given option in the velocity equation to compare it with others and find the maximum velocity.
OR To go more mathematically, one can differentiate the expression for v and equate it to zero. Solving this we will obtain the value of t at which v is an extremum. Now we know that one extremum is at t=0, and that gives zero velocity. We don’t want this trivial answer where v=0, hence the differentiation method explained earlier is going to give a maximum velocity.

Complete step-by-step solution
The velocity of the particle is given by \[v = 2t(3 - t)m{s^{ - 1}}\] which varies at different times. So, we can use the trial and error method which is trying out all the given options to find its corresponding velocity and then after comparing them with each other we will know at what time velocity is maximum. We can do the same to calculate the minimum velocity also.
Case A: At time t = 2 s
$
  v = 2 \times 2(3 - 2) \\
  v = 4 \times 1 \\
  v = 4m{s^{ - 1}} \\
  $
Case B: At time t = 3 s
$
  v = 2 \times 3(3 - 3) \\
  v = 6 \times 0 \\
  v = 0m{s^{ - 1}} \\
  $
Case C: At time t = $\dfrac{2}{3}$s
$
  v = 2 \times \dfrac{2}{3}\left( {3 - \dfrac{2}{3}} \right) \\
  v = \dfrac{4}{3} \times \dfrac{7}{3} \\
  v = 3.11m{s^{ - 1}} \\
  $
Case D: At time t = \[\dfrac{3}{2}\]s
$
  v = 2 \times \dfrac{3}{2}\left( {3 - \dfrac{3}{2}} \right) \\
  v = 3 \times \dfrac{3}{2} \\
  v = 4.5m{s^{ - 1}} \\
  $
On comparing all the above answers we get that the velocity of that particle is maximum at time t = \[\dfrac{3}{2}\]s.
Hence the correct option is D

Note Alternative method,
\[v = 2t(3 - t)m{s^{ - 1}}\]
On, differentiating you get,
$\dfrac{{dv}}{{dt}} = - 4t + 6$
v is max when $\dfrac{{dv}}{{dt}} = 0$ and at t = \[\dfrac{3}{2}\] s