1. Distance and Displacement:

Total length of path $(A C B)$ covered by the particle in a definite time interval is called distance. Displacement vector or displacement is the minimum distance $(A B)$ and directed from initial position to final position.

(i) $Distance \geq |Displacement|$

2. Displacement in Terms of Position Vector:

(i) From $Δ O A B, \vec{r} = {\vec{r}}_{B} - {\vec{r}}_{A}$

${\vec{r}}_{B} = x_{2} \hat{i} + y_{2} \hat{j} + z_{2} \hat{k}$ and ${\vec{r}}_{A} = x_{1} \hat{i} + y_{1} \hat{j} + z_{1} \hat{k}$

3. Average Velocity:

$v_{avg} = <v> = \frac{Total displacement}{Total time taken} = \frac{{\vec{r}}_{f} - {\vec{r}}_{i}}{Δ t}$

4. Average Speed:

(i) $Average speed = \frac{Total distance travelled}{Total time taken}$

(ii) $Average speed \geq |Average velocity|$

5. Instantaneous Velocity:

$\vec{v} = \frac{d \vec{r}}{d t}$

6. Average Acceleration:

${\vec{a}}_{avg} = \frac{Δ \vec{v}}{Δ t} = \frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{Δ t}$

7. Instantaneous Acceleration:

$\vec{a} = \frac{d \vec{v}}{d t}$

8. Graphs in Uniformly Accelerated Motion Along a Straight Line $(a \neq 0)$ :

(i) If $x$ is a quadratic polynomial in terms of $t$ , then $x - t$ graph is a parabola.

(ii) If $v$ is a linear polynomial in terms of $t$ , then $v - t$ graph is a straight line of slope $a$ .

(iii) $a - t$ graph is a horizontal line because $a$ is constant.

9. Maxima and Minima:

(i) $\frac{d y}{d x} = 0$ and $\frac{d^{2} y}{d x^{2}} < 0$ at maxima.

(ii) $\frac{d y}{d x} = 0$ and $\frac{d^{2} y}{d x^{2}} > 0$ at minima.

10. Equations of Motion (for Constant Acceleration):

(i) $v = u + a t$

(ii) $s = u t + \frac{1}{2} a t^{2}, x_{f} = x_{i} + u t + \frac{1}{2} a t^{2}$

(iii) $v^{2} = u^{2} + 2 a s$

(iv) $s = \frac{(u + v)}{2} t$

(v) $s_{n} = u + \frac{a}{2} (2 n - 1)$

11. Projectile Motion:

(i) A body crosses two points at same height in time $t_{1}$ and $t_{2}$ the points are at distance $x$ and $y$ from starting point,

(a) $x + y = R$

(b) $t_{1} + t_{2} = T$

(d) Average velocity from $A$ to $B$ is $u \cos (θ)$

(ii) If a person can throw a ball to a maximum distance $x$ , then the maximum height to which he can throw the ball will be $(\frac{x}{2})$ .

(iii) Velocity of particle at time $t$ ,

$\vec{v} = v_{x} \hat{i} + v_{y} \hat{j} = u_{x} \hat{i} + (u_{y} - g t) \hat{j} = u \cos (θ) \hat{i} + (u \sin (θ) - g t) \hat{j}$

(iv) If angle of velocity $\vec{v}$ from horizontal is $α$ , then

$\tan (α) = \frac{v_{y}}{v_{x}} = \frac{u_{y} - g t}{u_{x}} = \frac{u \sin (θ) - g t}{u \cos (θ)} = \tan (θ) - \frac{g t}{u \cos (θ)}$

(v) At highest point: $v_{y} = 0, v_{x} = u \cos (θ)$

(vi) Time of flight, $T = \frac{2 u_{y}}{g} = \frac{2 u \sin (θ)}{g}$

(vii) Horizontal range, $R = u \cos (θ) T = \frac{2 u^{2} \sin (θ) \cos (θ)}{g} = \frac{u^{2} \sin (2 θ)}{g} = \frac{2 u_{x} u_{y}}{g}$

Range is same for $θ$ and $(90 ° - θ)$ and maximum for $θ = 45 °$ .

(viii) Maximum height, $H = \frac{u_{y}^{2}}{2 g} = \frac{u^{2} \sin^{2} (θ)}{2 g} = \frac{1}{8} g T^{2}$ .

(ix) $\frac{H}{R} = \frac{1}{4} \tan (θ)$

(x) Equation of trajectory, $y = x \tan (θ) - \frac{g x^{2}}{2 u^{2} \cos^{2} (θ)} = x \tan (θ) (1 - \frac{x}{R})$

(xi) Horizontal projection from some height:

(a) Time of flight, $T = \sqrt{\frac{2 h}{g}}$ .

(b) Horizontal range, $R = u T = u \sqrt{\frac{2 h}{g}}$ .

(xii) Projectile motion on inclined plane (up motion):

(a) Time of flight, $T = \frac{2 u_{⊥}}{g_{⊥}} = \frac{2 u \sin (θ - α)}{g \cos (α)}$ .

(b) Maximum height, $H_{\max} = \frac{u_{⊥}^{2}}{2 g_{⊥}} = \frac{u^{2} \sin^{2} (θ - α)}{2 g \cos (α)}$ .

(d) Maximum range, $R_{\max} = \frac{u^{2}}{g (1 + \sin (α))}$ at angle, $θ = \frac{π}{4} + \frac{α}{2}$

(xiii) Projectile motion on inclined plane (down motion):

(a) Time of flight, $T = \frac{2 u_{⊥}}{g_{⊥}} = \frac{2 u \sin (θ + α)}{g \cos (α)}$ .

(b) Maximum height, $H_{\max} = \frac{u_{⊥}^{2}}{2 g_{⊥}} = \frac{u^{2} \sin^{2} (θ + α)}{2 g \cos (α)}$ .

(d) Maximum range, $R_{\max} = \frac{u^{2}}{g (1 - \sin (α))}$ at an angle, $θ = \frac{π}{4} - \frac{α}{2}$ .

12. Relative Motion:

(i) Velocity of $A$ with respect to $B$ , ${\vec{v}}_{AB} = {\vec{v}}_{A} - {\vec{v}}_{B}$ .

(ii) Acceleration of $A$ with respect to $B$ , ${\vec{a}}_{AB} = {\vec{a}}_{A} - {\vec{a}}_{B}$ .

(iii) Relative motion along a straight line, ${\vec{x}}_{BA} = {\vec{x}}_{B} - {\vec{x}}_{A}$ .

(iv) Shortest time in crossing river problems:

Let $v_{m}$ is the velocity of man with respect to ground, $v_{R}$ is the velocity of river and $v_{Rm}$ is the velocity of river with respect to man.

(a) Velocity along the river, $v_{x} = v_{R}$ .

(b) Velocity perpendicular to the river, $v_{y} = v_{mR} = \sqrt{v_{m}^{2} - v_{R}^{2}}$ .

(d) Time to cross the river, $T = \frac{d}{v_{mR}} = \frac{d}{\sqrt{v_{m}^{2} - v_{R}^{2}}}$ .

(e) Drift, $x = v_{R} T = v_{R} \frac{d}{\sqrt{v_{m}^{2} - v_{R}^{2}}}$ .

(v) Shortest path in crossing river problems:

(a) Velocity along the river, $v_{x} = 0$ .

(b) Velocity perpendicular to the river, $v_{y} = \sqrt{v_{mR}^{2} - v_{R}^{2}}$ .

(d) At an angle of $90 °$ with the river direction, velocity $v_{y}$ is used only to cross the river, therefore time to cross the river, $t = \frac{d}{v_{y}} = \frac{d}{\sqrt{v_{mR}^{2} - v_{R}^{2}}}$

(e) Velocity $v_{x}$ is zero, therefore, in this case the drift should be zero.

$v_{x} = 0$ or $v_{R} - v_{mR} \sin (θ) = 0$ or $v_{R} = v_{mR} \sin (θ)$ or $θ = \sin^{- 1} (\frac{v_{R}}{v_{mR}})$

(vi) Rain problems:

(a) Velocity of rain with respect to man, ${\vec{v}}_{Rm} = {\vec{v}}_{R} - {\vec{v}}_{m}$ .

(b) $v_{Rm} = \sqrt{v_{R}^{2} + v_{m}^{2}}$

13. For Circular Motion:

(i) Average angular velocity, $ω_{avg} = \frac{θ_{2} - θ_{1}}{t_{2} - t_{1}} = \frac{Δθ}{Δ t}$ .

(ii) Instantaneous angular velocity, $ω = \frac{d θ}{d t}$ .

(iii) Average angular acceleration, $α_{avg} = \frac{ω_{2} - ω_{1}}{t_{2} - t_{1}} = \frac{Δ ω}{Δθ}$ .

(iv) Instantaneous angular acceleration, $α = \frac{d ω}{d t} = ω \frac{d ω}{dθ}$ .

(v) Relation between speed and angular velocity, $v = r ω$ and $\vec{v} = \vec{ω} \times \vec{r}$

(vi) Tangential acceleration (rate of change of speed), $a_{t} = \frac{d v}{d t} = r \frac{d ω}{d t} = ω \frac{d r}{d t}$ .

(vii) Radial or normal or centripetal acceleration, $a_{r} = \frac{v^{2}}{r} = ω^{2} r$ .

(viii) Total acceleration, $\vec{a} = {\vec{a}}_{t} + {\vec{a}}_{r} \Rightarrow a = \sqrt{(a_{t}^{2} + a_{r}^{2})}$ .

(ix) ${\vec{a}}_{t} = \vec{a} \times \vec{r}$

(x) ${\vec{a}}_{r} = \vec{ω} \times \vec{v}$

(xi) Angular acceleration, $\vec{α} = \frac{d \vec{ω}}{d t}$ (non-uniform circular motion),.

(xii) Radius of curvature, $R = \frac{v^{2}}{a_{⊥}} = \frac{m v^{2}}{F_{⊥}}$ .

(a) If $y$ is a function of $x$ , i.e., $y = f (x)$ , then $R = \frac{{[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}}$ .

Which of the following graphs represents the speed v of a vertical projectile as a function of time t?

Important Questions on Kinematics

Learn and Prepare for any exam you want !

Which of the following graphs represents the speed $v$ of a vertical projectile as a function of time $t$ ?