Question

The position coordinates of a particle moving in $XY$–plane as a function of time t are$x=2t^2+6t+25m$ and $y=t^2+2t+2m$. The speed of particle at $t=10sec$, is

• A. $31m/s$
• B. $51m/s$
• C. $71m/s$
• D. $81m/s$

Answer 1

The correct option is B

$51m/s$

Step 1: Given Data:

Equations of positions are $x=2t^2+6t+25m$ and $y=t^2+2t+2m$

Time $t=10sec$

Step 2: Formula used:

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

$\left(V\right)=\frac{ds}{dt}$

Velocity in the x-direction is given as-

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

Velocity in the y-direction is given as-

$V_y=\frac{dy}{dt}$

Step 3: Calculation of the speed of the particle:

According to the obtained relationship, the velocity in the x-direction is,

$$\begin{gathered} x=2t^2+6t+25m \\ \frac{dx}{dt}=4t+6 \end{gathered}$$

The velocity in the x-direction at $t=10$ is,

$$\begin{gathered} {\left(\frac{dx}{dt}\right)}_{t=10}=4\times 10+6 \\ {V}_x=46 \end{gathered}$$

Similarly, the velocity in the y-direction is,

$$\begin{gathered} y=t^2+2t+2m \\ \frac{dy}{dt}=2t+2 \end{gathered}$$

The velocity in the y-direction at $t=10$ is,

$$\begin{gathered} {\left(\frac{dy}{dt}\right)}_{t=10}=2\times 10+2 \\ {V}_y=22 \end{gathered}$$

The resultant speed is given by-

$\begin{array}{l}\begin{array}{l}V=\sqrt{V_x^2+V_y^2}\end{array}\\ \begin{array}{l}=\sqrt{{46}^2+{22}^2}=51m/s\end{array}\end{array}$

Hence, option B is the correct answer.