Kinematics-1D

From the acceleration-time graph, acceleration increases linearly with time at a constant rate, so $\frac{da}{dt} = k$.
Integrating gives acceleration as a linear function of time: $a = kt$.
Since $a = \frac{dv}{dt}$, we have $\frac{dv}{dt} = kt$.
Integrating with respect to time gives $v = \int kt \, dt = \frac{1}{2} k t^2$.
This shows that velocity $v$ depends parabolically on time $t$, producing a curve that curves upwards.
When acceleration suddenly drops to zero, the rate of change of velocity becomes zero ($\frac{dv}{dt} = 0$), meaning velocity becomes constant.
Therefore, the graph consists of a parabolic curve followed by a horizontal straight line, which is represented by Graph (c).

For a physically real motion, a body cannot be at two different velocities at the exact same instant of time.
In Graph (d), drawing a vertical line parallel to the velocity axis intersects the graph at multiple points.
This indicates that the body possesses two different values of velocity at a single instant of time $t$.
Hence, Graph (d) is physically impossible and cannot represent real motion.

The ratio of applied forces during intervals AB and BC is equal to the ratio of acceleration during those intervals.
$\frac{F_{AB}}{F_{BC}} = \frac{a_{AB}}{a_{BC}}$
Acceleration during segment AB is given by the slope of AB, which makes an angle of $30^\circ$ with the positive time axis: $a_{AB} = \tan(30^\circ) = \frac{1}{\sqrt{3}}$.
Acceleration during segment BC is given by the slope of BC, which makes an angle of $120^\circ$ with the positive time axis: $a_{BC} = \tan(120^\circ) = -\sqrt{3}$.
Substitute these values into the ratio: $\frac{a_{AB}}{a_{BC}} = \frac{\frac{1}{\sqrt{3}}}{-\sqrt{3}}$.
Simplify the expression: $\frac{a_{AB}}{a_{BC}} = -\frac{1}{\sqrt{3} \times \sqrt{3}} = -\frac{1}{3}$.
Therefore, the ratio of applied forces for intervals AB and BC is $- \frac{1}{3}$.

Both cars start from rest at the same time, so their initial velocities are zero.
The slope of the velocity-time graph gives the acceleration of the body.
At any given instant of time $t$, the angle made by curve A with the time axis ($\theta_A$) is greater than the angle made by curve B ($\theta_B$).
Since $\theta_A > \theta_B$, it follows that $\tan \theta_A > \tan \theta_B$.
Therefore, the acceleration of car A is greater than the acceleration of car B.

Initially, when the ball is released from rest at a height, its velocity starts from zero and increases linearly in the negative downward direction as time progresses.
Just after rebounding from the floor, the direction of velocity immediately flips to positive (upward direction), and its magnitude is slightly smaller due to energy loss during collision.
As the ball rises upward against gravity, its positive velocity decreases linearly with time until it reaches zero at its peak height.
This process of bouncing and losing energy repeats continually until the ball eventually comes to rest.
This behavior of alternating linear downward and upward velocity profiles with decaying peak magnitudes is correctly represented in graph (a).

Acceleration along segment OA:
Change in velocity along OA is $v_2 - v_1 = 10 - 0 = 10\text{ m/s}$.
Time taken along OA is $t_2 - t_1 = 10 - 0 = 10\text{ s}$.
$a_{OA} = \frac{10 - 0}{10 - 0} = 1\text{ m/s}^2$.
Acceleration along segment AB:
Change in velocity along AB is $v_2 - v_1 = 10 - 10 = 0\text{ m/s}$.
Time taken along AB is $t_2 - t_1 = 20 - 10 = 10\text{ s}$.
$a_{AB} = \frac{10 - 10}{20 - 10} = 0\text{ m/s}^2$.
Acceleration along segment BC:
Change in velocity along BC is $v_2 - v_1 = 0 - 10 = -10\text{ m/s}$.
Time taken along BC is $t_2 - t_1 = 40 - 20 = 20\text{ s}$.
$a_{BC} = \frac{0 - 10}{40 - 20} = \frac{-10}{20} = -0.5\text{ m/s}^2$.
Therefore, the acceleration values along OA, AB, and BC are 1, 0, and -0.5 respectively.