Kinematics-1D

To determine the nature of the motion, calculate the average velocity for each 1-second interval:
For time interval $t = 0 \text{ s}$ to $t = 1 \text{ s}$:
$v_1 = \frac{0 - (-2)}{1 - 0} = 2 \text{ m/s}$
For time interval $t = 1 \text{ s}$ to $t = 2 \text{ s}$:
$v_2 = \frac{6 - 0}{2 - 1} = 6 \text{ m/s}$
For time interval $t = 2 \text{ s}$ to $t = 3 \text{ s}$:
$v_3 = \frac{16 - 6}{3 - 2} = 10 \text{ m/s}$
Now, calculate the rate of change of velocity (acceleration) between consecutive time intervals:
$a_1 = \frac{v_2 - v_1}{\Delta t} = \frac{6 - 2}{1} = 4 \text{ m/s}^2$
$a_2 = \frac{v_3 - v_2}{\Delta t} = \frac{10 - 6}{1} = 4 \text{ m/s}^2$
Since the acceleration remains constant at $4 \text{ m/s}^2$, the motion is uniform, accelerated.

For uniform motion, the speed of the body remains constant with time.
The relationship between distance $s$ and time $t$ for uniform motion is given by:
$s = v \cdot t$
Since velocity $v = \frac{ds}{dt}$ represents the slope of the $s-t$ graph, a constant velocity implies a constant slope.
A straight line inclined at a constant angle to the time axis represents a constant slope, and therefore, it correctly represents uniform motion.

The velocity of a particle from a displacement-time graph is given by the slope of the line:
$v = \tan\theta$
For particle $A$:
$v_A = \tan 30^\circ = \frac{1}{\sqrt{3}}$
For particle $B$:
$v_B = \tan 60^\circ = \sqrt{3}$
The ratio of their velocities $v_A : v_B$ is:
$\frac{v_A}{v_B} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{\sqrt{3} \times \sqrt{3}} = \frac{1}{3}$
Thus, the ratio $v_A : v_B$ is $1 : 3$.

In the given graph, time is plotted on the vertical axis and displacement is plotted on the horizontal axis.
The angle made by the line with the displacement axis is $30^\circ$.
Therefore, the angle made by the line with the time axis is:
$\theta = 90^\circ - 30^\circ = 60^\circ$
Velocity is given by the slope of the graph with respect to the time axis:
$v = \tan\theta = \tan 60^\circ = \sqrt{3} \text{ m/s}$

From the given displacement-time graph:
At initial time $t = 0 \text{ s}$, the initial position is $x(0) = 0 \text{ m}$.
At final time $t = 5 \text{ s}$, the final position is $x(5) = -10 \text{ m}$.
Total displacement $\Delta x = x(5) - x(0) = -10 - 0 = -10 \text{ m}$.
Total time interval $\Delta t = 5 - 0 = 5 \text{ s}$.
Average velocity is given by:
$v_{\text{av}} = \frac{\Delta x}{\Delta t} = \frac{-10}{5} = -2 \text{ ms}^{-1}$

From the given displacement-time graph:
1. Speed in the first second ($t = 0$ to $t = 1 \text{ s}$):
Displacement changes from $0 \text{ m}$ to $30 \text{ m}$ in $1 \text{ s}$.
$v_1 = \frac{30 - 0}{1 - 0} = 30 \text{ m/s}$
2. Speed in the next two seconds ($t = 1$ to $t = 3 \text{ s}$):
Displacement changes from $30 \text{ m}$ to $0 \text{ m}$ in $2 \text{ s}$.
$v_2 = \left\vert{} \frac{0 - 30}{3 - 1} \right\vert{} = \left\vert{} \frac{-30}{2} \right\vert{} = 15 \text{ m/s}$
3. Ratio of the speeds:
$\frac{v_1}{v_2} = \frac{30}{15} = \frac{2}{1}$
Thus, the ratio of the speed in the first second to that in the next two seconds is $2 : 1$.

During the upward motion, both gravity and air resistance act downwards, so the net retardation is $(g + a)$. Consequently, the speed decreases to zero rapidly with a larger slope magnitude.
During the downward motion, gravity acts downwards and air resistance acts upwards, so the net downward acceleration is $(g - a)$. Consequently, the speed increases from zero at a slower rate with a smaller slope magnitude.
Graph (c) correctly shows a steeper linear decrease in speed during the upward journey followed by a less steep linear increase in speed during the downward journey.

Maximum acceleration corresponds to the maximum positive slope in the speed-time graph.
Looking at the different segments of the graph:
1. From $t = 0 \text{ h}$ to $t = 0.25 \text{ h}$:
$a_1 = \frac{20 - 0}{0.25 - 0} = 80 \text{ km h}^{-2}$
2. From $t = 0.25 \text{ h}$ to $t = 0.75 \text{ h}$:
$a_2 = \frac{20 - 20}{0.75 - 0.25} = 0 \text{ km h}^{-2}$
3. From $t = 0.75 \text{ h}$ to $t = 1.00 \text{ h}$:
$a_3 = \frac{20 - 20}{1.00 - 0.75} = 0 \text{ km h}^{-2}$
4. From $t = 1.00 \text{ h}$ to $t = 1.25 \text{ h}$ (segment CD):
$a_{\text{max}} = \frac{60 - 20}{1.25 - 1.00} = \frac{40}{0.25} = 160 \text{ km h}^{-2}$
Therefore, the maximum acceleration during the journey is $160 \text{ km h}^{-2}$.

At $t = 0$, the tangent to the displacement-time curve has a positive slope, meaning the initial velocity is positive ($v > 0$). As time progresses towards the peak of the displacement curve, the slope becomes less steep, indicating that velocity decreases until it reaches zero ($v = 0$) at the maximum displacement. Past the peak, the slope becomes negative and increasingly steep, showing that the velocity becomes negative ($v < 0$). Since the rate of change of the slope is constant for a parabola, the velocity decreases linearly with time. Graph (a) correctly depicts a straight line starting from a positive velocity, crossing the time axis at $v = 0$, and continuing into negative velocity values.