Q. The position of a point in time ‘$t^{'}$ is given by $x = a + bt - ct^{2},y = at + bt^{2}.$Its acceleration at time ‘$t^{'}$is
A) $b - c$
B) $b + c$
C) 2$b - 2$c
D) $2\sqrt{b^{2} + c^{2}}$
Explanation
(d)
Given point is $x = a + bt - ct^{2}$
Acceleration in $x$ direction $= \frac{d^{2}x}{dt^{2}} = - 2c$
and acceleration in $y$ direction $= \frac{d^{2y}}{dt^{2}} = 2b$
$\therefore$ Resulatant acceleration
$= \sqrt{\left( \frac{d^{2x}}{dt^{2}} \right)^{2} + \left( \frac{d^{2y}}{dt^{2}} \right)^{2}} = \sqrt{( - 2c)^{2} + (2b)^{2}} = 2\sqrt{b^{2} + c^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.