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

The correct option is D $2\sqrt{b^2+c^2}$

Given,

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

Velocity along x-axis,

$v_x=\frac{dx}{dt}=b-2ct$

acceleration along x-axis,

$a_x=\frac{d{v}_x}{dt}=-2c$

Given,

$y=at+b{t}^2$

Velocity along y-axis,

$v_y=\frac{dy}{dt}=a+2bt$

Acceleration along y-axis,

$a_y=\frac{d{v}_y}{dt}=2b$

Acceleration, $\overrightarrow{a}=a_x\hat{i}+a_y\hat{y}$

$\overrightarrow{a}=-2c\hat{i}+2b\hat{j}$

$|\overrightarrow{a}|=a=\sqrt{(-2c{)}^2+(2b{)}^2}$

$a=\sqrt{4b^2+4c^2}$

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

The correct option is D.