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Motion in a Plane - Motion in a Plane with Constant Acceleration

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Motion in a plane with constant acceleration can be resolved into two independent one-dimensional motions along perpendicular axes, typically the xx and yy axes.

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If the acceleration vector a⃗\vec{a} is constant, its components axa_x and aya_y along the xx and yy directions are also constant.

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The velocity vector v⃗\vec{v} at any time tt is the vector sum of its initial velocity and the velocity gained due to acceleration: v⃗=v⃗0+a⃗t\vec{v} = \vec{v}_0 + \vec{a}t.

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The position vector r⃗\vec{r} at any time tt is calculated by integrating the velocity equation: r⃗=r⃗0+v⃗0t+12a⃗t2\vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2.

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The independence of the xx and yy components implies that changes in the yy-direction do not affect the motion in the xx-direction, and vice-versa, provided the axes are orthogonal.

📐Formulae

v⃗=v⃗0+a⃗t\vec{v} = \vec{v}_0 + \vec{a}t

r⃗=r⃗0+v⃗0t+12a⃗t2\vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2

vx=v0x+axtv_x = v_{0x} + a_x t

vy=v0y+aytv_y = v_{0y} + a_y t

x=x0+v0xt+12axt2x = x_0 + v_{0x} t + \frac{1}{2} a_x t^2

y=y0+v0yt+12ayt2y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2

v2=v02+2a⃗⋅(r⃗−r⃗0)v^2 = v_0^2 + 2 \vec{a} \cdot (\vec{r} - \vec{r}_0)

💡Examples

Problem 1:

A particle starts from the origin at t=0t = 0 with an initial velocity v⃗0=5.0i^ m/s\vec{v}_0 = 5.0 \hat{i} \text{ m/s} and moves in the x−yx-y plane with a constant acceleration a⃗=(3.0i^+2.0j^) m/s2\vec{a} = (3.0 \hat{i} + 2.0 \hat{j}) \text{ m/s}^2. (a) At what time is the xx-coordinate of the particle 84 m84 \text{ m}? (b) What is the yy-coordinate of the particle at that time?

Solution:

(a) For the xx-coordinate, we use: x=v0xt+12axt2x = v_{0x}t + \frac{1}{2} a_x t^2 Substituting the given values: 84=5.0t+12(3.0)t284 = 5.0t + \frac{1}{2}(3.0)t^2 1.5t2+5.0t−84=01.5t^2 + 5.0t - 84 = 0 Multiplying by 22 to simplify: 3t2+10t−168=03t^2 + 10t - 168 = 0 Using the quadratic formula t=−b±b2−4ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}: t=−10±102−4(3)(−168)2(3)t = \frac{-10 \pm \sqrt{10^2 - 4(3)(-168)}}{2(3)} Calculation of the discriminant: 100+20162116\begin{array}{r} 100 \\ + 2016 \\ \hline 2116 \end{array} t=−10±21166=−10±466t = \frac{-10 \pm \sqrt{2116}}{6} = \frac{-10 \pm 46}{6} Since t>0t > 0, we take the positive root: t=366=6 st = \frac{36}{6} = 6 \text{ s} (b) For the yy-coordinate, we use: y=v0yt+12ayt2y = v_{0y}t + \frac{1}{2} a_y t^2 Since v0y=0v_{0y} = 0: y=0(6)+12(2.0)(6)2=36 my = 0(6) + \frac{1}{2}(2.0)(6)^2 = 36 \text{ m}

Explanation:

To solve 2D motion problems, we split the motion into horizontal (xx) and vertical (yy) components. In part (a), we solve for time tt using the displacement equation for the xx-axis. In part (b), we substitute this time value into the displacement equation for the yy-axis to find the vertical position.