MCQ
The displacement of a particle is given by $y = a + bt + c{t^2} - d{t^4}$. The initial velocity and acceleration are respectively
- A$b,\, - 4d$
- B$ - b,\,2c$
- ✓$b,\,2c$
- D$2c,\, - 4d$
✓
Answer
Correct option: C.
$b,\,2c$
c
(c) $y = a + bt + c{t^2} - d{t^4}$
(c) $y = a + bt + c{t^2} - d{t^4}$
$\therefore \;v = \frac{{dy}}{{dt}} = b + 2ct - 4d{t^3}$ and $a = \frac{{dv}}{{dt}} = 2c - 12d{t^2}$
Hence, at $t = 0$, $v_{initial} \,= b$ and $a_{initial}$ $= 2c$.
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