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

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

Step 1: Given Data:

Equations$\Rightarrow x=a+bt–c{t}^2,y=at+b{t}^2$

Time$=t$

Step 2: Formula used:

We know that the rate of change of displacement with respect to time is known as velocity. If the velocity is $V$ then,

$V=\frac{dx}{dt}$

And the rate of change of velocity with respect to time is known as acceleration. If the acceleration is $\alpha$ then,

$\alpha =\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}$

Therefore, the acceleration relationship with displacement is $a=\frac{d^{2}x}{d{t}^2}$

Step 3: Calculation of acceleration:

Acceleration in the x-direction,

$$\begin{gathered} x=a+bt-c{t}^2 \\ \frac{dx}{dt}=0+b-2tc \\ \frac{d^{2}x}{d{t}^2}=-2c \\ {\alpha }_x=-2c \end{gathered}$$

Acceleration in the y-direction,

$$\begin{gathered} y=at+b{t}^2 \\ \frac{dy}{dt}=a+2bt \\ \frac{d^{2}y}{d{t}^2}=2b \\ {\alpha }_y=2b \end{gathered}$$

Resultant acceleration,

$$\begin{gathered} a=\sqrt{{\left({\alpha }_x\right)}^2+{\left({\alpha }_y\right)}^2} \\ a=\sqrt{{(-2c)}^2+{\left(2b\right)}^2} \\ a=2\sqrt{b^2+c^2} \end{gathered}$$

Hence, option D is the correct answer.