Q109 / 155 · WBJEE 2024 · Mathematics

The acceleration $f \mathrm{ft} / \mathrm{sec}^2$ of a particle after a time $t$ sec starting from rest is given by $f=6-\sqrt{1.2 t}$. Then the maximum velocity v and time T at attend this velocity are

  • A
    T = 20 sec
  • Correct answer
    B
    v = 60 ft/sec
  • Correct answer
    C
    T = 30 sec
  • D
    v = 40 tt/sec

Solution

Hint: Given $a^n c^n=f=\frac{d v}{d t}=6-\sqrt{1.2 t} \quad(t$ is in secs) at $\mathrm{t}=0, \mathrm{v}=0$ and let the maximum velocity is ' v ' which is attained at ' $\mathrm{t}_0$ ' at maximum velocity, $\frac{\mathrm{dv}}{\mathrm{dt}}=0 \Rightarrow \mathrm{f}=0 \Rightarrow \mathrm{t}_0=30 \mathrm{~s}$ so, $\int_0^v d v=\int_0^{30}(6-\sqrt{1.2 t}) d t \Rightarrow v=60 \mathrm{ft/s}$

Hence, the correct answer is options: 'B', 'C'.

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