Kinematics-1D

A → (s, u)
$\mathrm{B} \rightarrow (\mathrm{s}, \mathrm{t})$
$C \to (s, u)$
$\mathrm{D} \rightarrow (\mathrm{s}, t)$
$\mathrm{E} \rightarrow (\mathrm{q}, \mathrm{r})$
$\mathrm{F} \rightarrow (\mathrm{s}, \mathrm{u})$
$G \to (p, s)$
Between t=0 and t=1 s, motion is decelerated $a\neq0$ .
$(\mathrm{A}) \rightarrow (\mathrm{s}, \mathrm{u})$
Between t=1 s and t=2 s, motion is accelerated $a\neq0$ .
$\mathrm{So}, (\mathrm{B}) \rightarrow (\mathrm{s}, \mathrm{t})$
Between t = 2 s and t = 3 s, motion is decelerated ( $a \neq 0$ ).
$\mathrm{So}, (\mathrm{C}) \rightarrow (\mathrm{s}, \mathrm{u})$
Between $t = 3 \, \text{s}$ and $t = 4 \, \text{s}$ , motion is accelerated $a \neq 0$ .
$\mathrm{So}, (\mathrm{D}) \rightarrow (\mathrm{s}, \mathrm{t})$
Between $t = 4 \, \text{s}$ and $t = 5 \, \text{s}$ , motion is uniform i.e., $v \neq 0$ and $a = 0$ .
$\mathrm{So}, (\mathrm{E}) \rightarrow (\mathrm{q}, \mathrm{r})$
Between $t = 5 \, \text{s}$ and $t = 6 \, \text{s}$ , motion is decelerated ( $a \neq 0$ ).
$\mathrm{So}, (\mathrm{F}) \rightarrow (\mathrm{s}, \mathrm{u})$
At t=1 s and at t=3 s the motion reverses its direction, so v=0 and simultaneously $a\neq0$ .
$\mathrm{So}, (\mathrm{G}) \rightarrow (\mathrm{p}, \mathrm{s})$

A → (q)
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \to (\mathrm{s})$
$\mathrm{D} \rightarrow (\mathrm{p})$
$x = - 5 t ^ {2} + 2 0 t + 1 0$
$\Rightarrow \frac {d x}{d t} = - 1 0 t + 2 0$
$\Rightarrow v = 2 0 - 1 0 t\tag{... (1}$
Now v=0 at t=2 s, so the distance travelled by the particle from t=0 to t=3 s is
$x = \left| \int_ {0} ^ {2} v d t \right| + \left| \int_ {2} ^ {3} v d t \right|$
$\mathrm{Now} \int v d t = (2 0 - 1 0 t) d t$
$\Rightarrow \int v d t = 2 0 t - 5 t ^ {2}$
$\Rightarrow x = \left| (2 0 t - 5 t ^ {2}) \right| _ {0} ^ {2} + \left| (2 0 t - 5 t ^ {2}) \right| _ {2} ^ {3}$
$\Rightarrow x = \vert 4 0 - 2 0 \vert + \vert 8 0 - 8 0 - 4 0 + 2 0 \vert$
$\Rightarrow x = | 2 0 | + | - 2 0 |$
$\Rightarrow x = 4 0 \mathrm{m}$
So, average speed is
$v _ {a v} = \frac {x}{t} = \frac {4 0}{4} = 1 0 \mathrm{ms} ^ {- 1}$
Displacement of the particle from t = 0 to t = 4 s is
$\Delta x = \int_ {0} ^ {4} v d t = \int_ {0} ^ {4} (2 0 - 1 0 t) d t$
$\Rightarrow \Delta x = (2 0 t - 5 t ^ {2}) \big | _ {0} ^ {4}$
$\Rightarrow \Delta x = 8 0 - 8 0 = 0$
So, average velocity is
$v _ {a v} = \frac {\Delta x}{\Delta t} = \frac {0}{4} = 0 \mathrm{ms} ^ {- 1}$
From (1), $a = \frac{dv}{dt} = -10\mathrm{ms}^{-2}$
$u = v | _ {t = 0} = 2 0 \mathrm{ms} ^ {- 1}$
Since, initially $u = \oplus$ and $a = \ominus$ , so a happens to be retardation for the motion from t = 0 to t = 2 s (till its velocity becomes zero momentarily i.e., the particle reverses its direction of motion).
Velocity at $t = 4$ s is
$v | _ {t = 4 \mathrm{s}} = 2 0 - 1 0 (4) = - 2 0 \mathrm{ms} ^ {- 1}$
Hence, speed at t = 4 s is $20 \, ms^{-1}$

A $\rightarrow$ (q)
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \to (p, s)$
$\mathrm{D} \rightarrow (t)$
$\frac {d \vec {r}}{d t} = \mathrm{Velocity}$
$\frac {d | \vec {v} |}{d t} = \text { Tangential Acceleration }$
$\frac {d \vec {v}}{d t} = \frac {d ^ {2} \vec {r}}{d t ^ {2}} = \mathrm{Acceleration}$
$\left| \frac {d \vec {r}}{d t} \right| = \text { Instantaneous speed }$

A $\rightarrow (\mathbf{r},\mathbf{t})$
$\mathrm{B} \rightarrow (\mathrm{p})$
$\mathrm{C} \rightarrow (\mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{s})$
$x = (t - 3) ^ {2}$
$\Rightarrow x = t ^ {2} - 6 t + 9$
$\Rightarrow v = \frac {d x}{d t} = 2 t - 6$
2 t - 6 = 0
$\Rightarrow t = 3 \mathrm{s}$

t	0	1	2	3	4	5	6
x	9	4	1	0	1	4	9
v	-6	-4	-2	0	2	4	6
a	2	2	2	2	2	2	2
Nature of Motion	Decelerating	Decelerating	Decelerating	Accelerating	Accelerating	Accelerating	Accelerating

Displacement of the particle from t=0 to t=6 s is zero.
Distance travelled by the particle from t = 0 to t = 6 is $9 + 9 = 18 \, m$
Average speed of the particle is
$v _ {a v} = \frac {\text { Total Distance Travelled }}{\text { Total Time Taken }}$
$\Rightarrow v _ {a v} = \frac {1 8}{6} = 3 \mathrm{ms} ^ {- 1}$
Average Velocity of the particle is
$\left| \vec {v} _ {a v} \right| = \frac {\text { Displacement }}{\text { Time }} = 0$
Acceleration of the particle over the entire duration of motion is $2 \, ms^{-2}$ .

A → (t)
$\mathrm{B} \rightarrow (\mathrm{s})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\mathrm{q})$
$\mathrm{E} \rightarrow (\mathrm{r})$
Given, that $v_{br} = 4 \, kmhr^{-1}$ and $v_{r} = 2 \, kmhr^{-1}$
$\Rightarrow \quad \alpha = \sin^ {- 1} \left(\frac {v _ {r}}{v _ {b r}}\right) = \sin^ {- 1} \left(\frac {2}{4}\right) = \sin^ {- 1} \left(\frac {1}{2}\right) = 3 0 ^ {\circ}$
Hence, to reach the point directly opposite to starting point he should head the boat at an angle of $30^{\circ}$ with AB or $90^{\circ} + 30^{\circ} = 120^{\circ}$ with the river flow.

Time taken by the boatman to cross the river for zero drift condition is
$t _ {1} = \frac {l}{v _ {b r} \cos \alpha}$
where $l = 4 \, km$ , $v_{br} = 4 \, kmh^{-1}$ and $\alpha = 30^{\circ}$
$\Rightarrow t _ {1} = \frac {4}{4 \cos (3 0 ^ {\circ})} = \frac {2}{\sqrt {3}} \mathrm{hr}$
$\Rightarrow t_{1} = \frac{120}{\sqrt{3}} = 40\sqrt{3}$ minute
For shortest time $\theta=0^{\circ}$ i.e., the man must row the boat perpendicular to the river flow. So $\beta=90^{\circ}$
and $t_{\min}=t_{2}=\frac{l}{v_{br}\cos(0^{\circ})}=\frac{4}{4}=1\mathrm{hr}=60\mathrm{minute}$

Hence, he should head his boat perpendicular to the river current for crossing the river in shortest time and this shortest time $t_{2}$ of 60 minute.
$t = t _ {P \to Q} + t _ {Q \to P}$
$\Rightarrow t = \frac {P Q}{v _ {b r} + v _ {r}} + \frac {Q P}{v _ {b r} + v _ {r}}$
$\Rightarrow t = \frac {2}{4 + 2} + \frac {2}{4 - 2} = \frac {4}{3} \mathrm{hr}$
$\Rightarrow$ t = 80 minute

A → (q) B → (r) C → (p, q) D → (s)

A → (r) B → (p) C → (q) D → (p)

A → (s) B → (r) C → (p) D → (q)

$\begin{array}{l} \mathrm{A} \to (\mathrm{r}) \\ \mathrm{B} \to (\mathrm{p}) \\ \mathrm{C} \to (\mathrm{r}) \\ \mathrm{D} \to (\mathrm{s}) \end{array}$
$v _ {i} = + 1 0 \mathrm{ms} ^ {- 1} \text {and} v _ {f} = 0$
$\Rightarrow \Delta v = v _ {f} - v _ {f} = - 1 0 \mathrm{ms} ^ {- 1}$
$\Rightarrow a _ {a v} = \frac {\Delta v}{\Delta t} = \frac {- 1 0}{6} = \frac {- 5}{3} \mathrm{ms} ^ {- 2}$
Total displacement = area under v-t graph (with sign) and acceleration = slope of v-t graph.

A → (r)
$\begin{array}{l} \mathrm{A} \to (\mathrm{r}) \\ \mathrm{B} \to (\mathrm{p}) \\ \mathrm{C} \to (\mathrm{s}) \\ \mathrm{D} \to (\mathrm{q}) \end{array}$
After 2 s velocity of balloon and hence the velocity of the particle will be $20 \, \text{ms}^{-1} (= \text{at})$ and its height from the ground will be $20 \, m \left( = \frac{1}{2} at^{2} \right)$ . Now g, will start acting on the particle.

A → (p)
$\mathrm{B} \rightarrow (\mathrm{q})$
$C \to (s)$
$\mathrm{D} \rightarrow (\mathrm{r})$
$\frac {v _ {m}}{t _ {1}} = \alpha \mathrm{and} \frac {v _ {m}}{t _ {2}} = \beta$
$\text { Also } x + y = \frac {1}{2} \times v _ {m} \times \left(t _ {1} + t _ {2}\right)$

$\Rightarrow \quad (x + y) = \frac {1}{2} \times v _ {m} \times \left(\frac {v _ {m}}{\alpha} + \frac {v _ {m}}{\beta}\right)$
$\Rightarrow v _ {m} = \sqrt {\frac {2 (x + y) \alpha \beta}{(\alpha + \beta)}}$
Now, $v_{av}=\frac{x+y}{t_{1}+t_{2}}=\frac{x+y}{\frac{v_{m}}{\alpha}+\frac{v_{m}}{\beta}}=\frac{\alpha\beta}{\alpha+\beta}\cdot\frac{x+y}{v_{m}}$

A → (q)
B → (p, q)
C → (p, r)
D → (r, s) $v = \frac{dx}{dt}$ $\Rightarrow a = \frac{dv}{dt}$ $\Rightarrow v = \int_{0}^{t} ad t$ $\Rightarrow x = \int_{0}^{t} v dt$

A → (r) B → (p) C → (r) D → (r)

A → (r)
B → (q)
C → (s)
D → (p)
A → v = Kt
$⇒ a = \frac{dv}{dt} = K$
$B → v = Kt^2$
$⇒ a = \frac{dv}{dt} = 2Kt$
$C → K_1t → First part of graph$
$⇒ a = \frac{dv}{dt} = K_1$
$⇒ v = K_2 → Second part of graph$
$⇒ a = \frac{dv}{dt} = 0$
$D → v = v_0e^{-Kt}$
$⇒ a = \frac{dv}{dt} = (-v_0K)e^{-Kt}$
$⇒ a = -a_0e^{-Kt}$
$⇒ a_0 = v_0K$

A → (r)
B → (s)
C → (p)
D → (q)
In motion M : slope of s-t graph is positive and increasing. Therefore, velocity of the particle is positive and increasing. Hence, it is A type motion. Similarly, N, P and Q can be observed from the slope.

A → (r) B → (q) C → (s) D → (p)

A → (r, s) B → (r, s) C → (p, q, r, t) D → (p, q, r, t)

A → (q, s) B → (r) C → (p) D → (s)

A → (r) B → (s) C → (p) D → (q)

A → (s)
$\mathrm{C} \rightarrow (\mathrm{q})$
Acceleration is constant and is equal to acceleration due to gravity, which is acting vertically downwards i.e. in the positive direction, so (p) represents the a-t graph for the motion.
Since $v = u + at$ and $s = \Delta y = ut + \frac{1}{2}gt^{2}$

Before collision, v = gt {∵ u = 0}
So, before collision velocity is increasing linearly with time and is increasing. However, after collision velocity decreases with time linearly and is negative as in (r).
$y = \frac {1}{2} g t ^ {2}$
tive, whereas after collision, the displacement is $y = -v_{0}t + \frac{1}{2}gt^{2}$ and is positive (a parabola opening
(upwards). Here is the tricky part, as we have taken the origin at the point from where the ball is dropped and in both the cases of upward and downward motion, the displacement is downwards and hence displacement is positive in both the cases.
The distance-time graph is positive and always increasing.

A → (p, r) B → (p, r) C → (p, q) D → (p, q)

A → (p, q)
$\mathrm{B} \rightarrow (\mathrm{p}, \mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{p}, \mathrm{q})$
With constant positive acceleration, speed increases, when velocity is positive and speed decreases, when velocity is negative.
Similarly, with constant negative acceleration speed increases, when velocity is negative and speed decreases, when velocity is positive.

A → (q)
B → (r)
C → (s)
D → (p)
Solve using relative velocity diagram.