WBJEE · Maths · Differentiation
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
- B v = 60 ft/sec
- C T = 30 sec
- D v = 40 tt/sec
Answer & Solution
Correct Answer
(C) T = 30 sec
Step-by-step Solution
Detailed explanation
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,…
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