Question

The position of a point in time 't' is given by $x=a+bt-c{t}^2,y=at+b{t}^2$. Its acceleration at time 't' is.

• A. $b-c$
• B. $b+c$
• C. $2b-2c$
• D. $2\sqrt{b^2+c^2}$

Answer 1

Answer: (D)

$2\sqrt{b^2+c^2}$

Given point is $x=a+bt-c{t}^2$

Acceleration in $x$ direction $=\frac{d^{2}x}{d{t}^2}=-2c$

and acceleration in y direction $=\frac{d^{2}y}{d{t}^2}=2b$

$\therefore$ Resultant acceleration is $\sqrt{[{\left(\frac{d^{2}x}{d{t}^2}\right)}^2+{\left(\frac{d^{2}y}{d{t}^2}\right)}^2]}=\sqrt{(-2c{)}^2+(2b{)}^2}=2\sqrt{b^2+c^2}$