Kinematics 2-D flash cards

1. 1.In 2-D motion, how are position, velocity and acceleration related as vectors?

   Position $\vec r = x\hat i + y\hat j$. Velocity $\vec v = \dfrac{d\vec r}{dt} = v_x\hat i + v_y\hat j$. Acceleration $\vec a = \dfrac{d\vec v}{dt} = a_x\hat i + a_y\hat j$. Each component evolves independently.

   Hint: Differentiate component-wise.

2. 2.What is the key principle that lets us treat 2-D motion as two 1-D problems?

   The independence of perpendicular components: motion along $x$ and along $y$ are governed by separate equations sharing only the common variable time $t$. Horizontal and vertical motions do not affect each other.

   Hint: $x$ and $y$ share only $t$.

3. 3.Write the vector equations of motion for constant acceleration $\vec a$ in a plane.

   $\vec v = \vec u + \vec a t$ and $\vec r = \vec u t + \tfrac12 \vec a t^2$. These hold component-wise: $x = u_x t + \tfrac12 a_x t^2$, $y = u_y t + \tfrac12 a_y t^2$.

   Hint: Same as 1-D but vectorial.

4. 4.How do you find the magnitude and direction of a resultant vector $\vec A = A_x\hat i + A_y\hat j$?

   Magnitude $|\vec A| = \sqrt{A_x^2 + A_y^2}$; direction $\theta = \tan^{-1}\!\left(\dfrac{A_y}{A_x}\right)$ measured from the $+x$ axis (mind the quadrant).

   Hint: Pythagoras + arctan.

5. 5.What is the average velocity vector over an interval, and how does it differ from average speed?

   $\vec v_{avg} = \dfrac{\Delta \vec r}{\Delta t}$ (displacement over time, a vector). Average speed $= \dfrac{\text{total path length}}{\Delta t}$ (a scalar $\ge |\vec v_{avg}|$).

   Hint: Displacement vs path length.

6. 6.For any projectile launched with speed $u$ at angle $\theta$ (ground level), state the velocity components at launch.

   $u_x = u\cos\theta$ (constant throughout), $u_y = u\sin\theta$ (changes under gravity). Acceleration $a_x = 0$, $a_y = -g$.

   Hint: Horizontal is uniform.

7. 7.Derive the time of flight $T$ for a projectile launched at angle $\theta$ from level ground.

   Vertical: $y = u\sin\theta\, t - \tfrac12 g t^2 = 0$ at landing. So $t(u\sin\theta - \tfrac12 g t)=0 \Rightarrow T = \dfrac{2u\sin\theta}{g}$.

   Hint: Set $y=0$, take nonzero root.

8. 8.Derive the maximum height $H$ of a projectile launched at angle $\theta$.

   At the top $v_y = 0$: $0 = (u\sin\theta)^2 - 2gH \Rightarrow H = \dfrac{u^2\sin^2\theta}{2g}$. It is reached at $t = \dfrac{u\sin\theta}{g} = T/2$.

   Hint: Use $v_y^2 = u_y^2 - 2gH$.

9. 9.Derive the horizontal range $R$ of a projectile on level ground.

   $R = u\cos\theta \cdot T = u\cos\theta \cdot \dfrac{2u\sin\theta}{g} = \dfrac{u^2\sin 2\theta}{g}$.

   Hint: $R = u_x T$; use $2\sin\theta\cos\theta=\sin2\theta$.

10. 10.At what launch angle is the range maximum, and what is $R_{max}$?

    $R = \dfrac{u^2\sin2\theta}{g}$ is maximum when $\sin2\theta=1$, i.e. $\theta = 45^\circ$. Then $R_{max} = \dfrac{u^2}{g}$.

    Hint: $\sin2\theta=1$.

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