Kinematics-1D

$x=\frac{1}{2}at^{2}$ (if coefficients of $t^{2}$ is positive (a >
0) upward opening parabola)

S/2 S/2
υ 2υ
$V _ {\text {avg}} = \frac {2 v _ {1} v _ {2}}{v _ {1} + v _ {2}} = \frac {2 (v) (2 v)}{v + 2 v} = \frac {4 v ^ {2}}{3 v} = \frac {4 v}{3}$

$S = \mathrm{ut} + \frac{1}{2}\mathrm{at}^2$

$- \mathrm{H} = 4 \times 4 - \frac {1}{2} \times 1 0 \times 4 ^ {2}$
$- \mathrm{H} = 1 6 - 8 0$
$- \mathrm{H} = - 6 4$
$\mathrm{H} = 6 4 \mathrm{m}$

By $v^{2} = u^{2} + 2$ as
$\left(\frac {u}{3}\right) ^ {2} = u ^ {2} - 2 a x$
$2 \mathrm{ax} = \mathrm{u} ^ {2} - \frac {\mathrm{u} ^ {2}}{9}$
$\boxed {2 a x = \frac {8 u ^ {2}}{9}}$
... (1)
Similarly from starting
$\mathbf {v} ^ {2} = \mathbf {u} ^ {2} + 2 \mathbf {a x}$
$0 = \mathrm{u} ^ {2} - 2 \mathrm{ax} _ {2}$
$\boxed {2 \mathrm{ax} _ {2} = \mathrm{u} ^ {2}}$
... (2)
$\mathrm{By(1)} / (2)$
$\frac {\mathrm{x}}{\mathrm{x} _ {2}} = \frac {8}{9}$
$\frac {2 4}{x _ {2}} = \frac {8}{9}$
$\mathrm{x} _ {2} = 2 7 \mathrm{cm}$

$\vec{V}=\frac{d\vec{r}}{dt}=4\hat{i}+4t\hat{j}+0\hat{k}$
$\mathrm{att=1sec}$
$\vec {\mathrm{V}} = 4 \hat {\mathrm{i}} + 4 (1) \hat {\mathrm{j}}$
$| \vec {V} | = \sqrt {4 ^ {2} + 4 ^ {2}} = 4 \sqrt {2}$
$\tan \alpha = \frac {4}{4} = 1$
$\alpha = 4 5 ^ {\circ}$

At the top most point of its trajectory particle will have only horizontal component of velocity
V at top = v cos θ
= 20 × $\frac{1}{2}$ = 10 m/s

$\begin{array}{l}\mathrm{x} = \alpha \mathrm{t}^4 +\beta \mathrm{t}^2 +\gamma \mathrm{t} + \delta \\ \mathrm{V} = \frac{\mathrm{dx}}{\mathrm{dt}} = 4\alpha \mathrm{t}^3 +2\beta \mathrm{t} + \gamma \\ \Rightarrow \mathrm{at}\qquad \mathrm{t} = 0,\qquad \mathrm{v} = \gamma \\ \mathrm{a} = \frac{\mathrm{dv}}{\mathrm{dt}} = 12\alpha \mathrm{t}^2 +2\beta \\ \Rightarrow \mathrm{at}\qquad \mathrm{t} = 0,\qquad \mathrm{a} = 2\beta \\ \therefore \frac{\text{Initial velocity}}{\text{Initial acceleration}} = \frac{\gamma}{2\beta} \end{array}$

D

$\left(\vec{V}_{i}\right)=(u\cos\theta)\hat{i}+(u\sin\theta)\hat{j}$ $(\vec{V}_{f})=(u\cos\theta)\hat{i}-(u\sin\theta)\hat{j}$ $(\vec{V}_{top})=(u\cos\theta)\hat{i}$ $|\vec{V}_{i}|=Initial speed=u$
$| \vec {V} _ {\mathrm{f}} | = \text { final speed } = u$
$\left| \vec {V} _ {\text { top }} \right| = \text { speed at top } = u \cos \theta (A)$
(B) $\overrightarrow{\Delta V} = \overrightarrow{V}_{\mathrm{f}} - \overrightarrow{V}_{\mathrm{i}}\& \Delta |\vec{V}| = |\vec{V}_{\mathrm{f}}| - |\vec{V}_{\mathrm{i}}| = 0$
$| \overrightarrow {\Delta V} | = (2 u \sin \theta) \hat {j}$
$| \overrightarrow {\Delta V} | = 2 u \sin \theta$
(C) $|\vec{\mathrm{V}}_{\mathrm{i}}| - |\vec{\mathrm{V}}_{\mathrm{top}}| = \mathrm{u} - \mathrm{u}\cos \theta$
(D) $|\vec{V}_i - \vec{V}_{\mathrm{top}}| = u\sin \theta$

$\begin{array}{l l} \vec {\mathrm{v}} _ {\mathrm{i}} = 3 \hat {\mathrm{i}} + 4 \hat {\mathrm{j}} & \Rightarrow | \vec {\mathrm{v}} _ {\mathrm{i}} | = 5 \\ \vec {\mathrm{v}} _ {\mathrm{f}} = 3 \hat {\mathrm{i}} + 4 \hat {\mathrm{j}} & \Rightarrow | \vec {\mathrm{v}} _ {\mathrm{f}} | = 5 \end{array}$
$\begin{array}{r l} \text {(A)} & | \Delta \vec {v} | = | \vec {v} _ {\mathrm{f}} - \vec {v} _ {\mathrm{i}} | = | 3 \hat {\mathrm{i}} + 4 \hat {\mathrm{j}} - 3 \hat {\mathrm{i}} + 4 \hat {\mathrm{j}} | \\ & = | 8 \hat {\mathrm{j}} | = 8 \end{array}$
(B) $\Delta |\vec{\mathrm{v}}| = |\vec{\mathrm{v}}_{\mathrm{f}}| - |\vec{\mathrm{v}}_{\mathrm{i}}| = 5 - 5 = 0$
(C) $\vec{v}_{f}-\vec{v}_{i}=8\hat{j}$
(D) $\vec{\mathbf{v}}_{\mathrm{f}} + \vec{\mathbf{v}}_{\mathrm{i}} = 6\hat{\mathbf{i}}$

from the figure, it is clear that time of flight is $(\mathbf{t}_1 + \mathbf{t}_2)$
for maximum Ht :-
time taken from B to C is $\left(\frac{t_1 + t_2}{2}\right)$
Applying $2^{nd}$ eq $^{n}$ of motion from B to C
$\mathrm{H} _ {\max} = 0 + \frac {1}{2} \mathrm{g} \left(\frac {\mathrm{t} _ {1} + \mathrm{t} _ {2}}{2}\right) ^ {2}$
for (u) $\Rightarrow$ Speed at A = Speed at C
Apply $1^{\mathrm{st}}$ eq of motion between B & C
$\mathbf {u} = \mathbf {0} + \mathbf {g} \left(\frac {\mathbf {t} _ {1} + \mathbf {t} _ {2}}{2}\right)$
$\text { for } (h) \Rightarrow h = u t _ {1} - \frac {1}{2} g t _ {1} ^ {2}$
$\mathbf {h} = \mathbf {g} \left(\frac {\mathbf {t} _ {1} + \mathbf {t} _ {2}}{2}\right) \mathbf {t} _ {1} - \frac {1}{2} \mathbf {g} \mathbf {t} _ {1} ^ {2}$
$\boxed {h = \frac {1}{2} g t _ {1} t _ {2}}$

Two projectiles having same range may have different time of flight.

Projectile motion is an example of motion under constant acceleration.

C

B

D

When body moves with constant velocity then it moves with constant speed in same direction,
So $\left|Inst. velocity\right| = \left|Avg. vel.\right|$ = Inst. speed = Avg speed

Average speed for first 6 s
$\begin{array}{l} = \frac {\text { distance travelled in } 6 \mathrm{s}}{6 \mathrm{s}} \\ = \frac {\frac {1}{2} \times 1 5 \times 4 + \frac {1}{2} \times 2 \times 1 0}{6} = \frac {2 0}{3} \mathrm{ms} ^ {- 1} \end{array}$

$\begin{array}{r l} 1 1. & \text { Average velocity for first } 1 2 \mathrm{s} \\ & = \frac {\text { displacement in first } 1 2 \mathrm{s}}{1 2 \mathrm{s}} \\ & = \frac {- \frac {1}{2} \times 1 5 \times 4 + \frac {1}{2} \times 2 \times 1 0 + 2 \times 1 0 + \left(\frac {2 0 + 1 0}{2}\right) (1 2 - 8)}{1 2} \\ & = \frac {6 0}{1 2} = 5 \mathrm{ms} ^ {- 1} \end{array}$

Average acceleration from t = 5 s to t = 15 s
$= \frac {\mathrm{v} (t = 1 5) - \mathrm{v} (t = 5)}{1 5 - 5} = \frac {0 - 5}{1 0} = - \frac {1}{2} \mathrm{ms} ^ {- 2}$

A