Motion in 1D

Step 1: Recall that the slope of a displacement-time ($s-t$) graph gives the velocity of the moving body.
Step 2: The slope is calculated as the tangent of the angle the graph makes with the time axis (positive x-axis).
Step 3: The problem implies the graph makes an angle of $30^\circ$ with the displacement axis (y-axis).
Step 4: Therefore, the angle made with the time axis is $90^\circ - 30^\circ = 60^\circ$.
Step 5: Calculate the velocity using the tangent of this angle: $v = \tan(60^\circ)$.
Step 6: The value of $\tan(60^\circ)$ is $\sqrt{3}$.
Step 7: Thus, the velocity of the moving body is $\sqrt{3} \mathrm{ms}^{-1}$.

Step 1: Recall that the slope of a distance-time graph represents the velocity of the particle.
Step 2: The initial velocity $v_i$ is the slope at time $t$, which is given by the tangent of the angle with the time axis.
Step 3: Calculate $v_i = \tan(45^\circ) = 1 \text{ ms}^{-1}$.
Step 4: The final velocity $v_f$ after one second is the new slope, given by the tangent of the new angle.
Step 5: Calculate $v_f = \tan(60^\circ) = \sqrt{3} \text{ ms}^{-1}$.
Step 6: The time interval $\Delta t$ is given as $1 \text{ s}$.
Step 7: Average acceleration $a_{av}$ is the change in velocity divided by the time interval.
Step 8: Calculate $a_{av} = \frac{v_f - v_i}{\Delta t} = \frac{\sqrt{3} - 1}{1} = \sqrt{3} - 1 \text{ ms}^{-2}$.

Step 1: Recall that the acceleration of an object is equal to the slope of its velocity-time (v-t) graph, calculated as $a = \frac{\Delta v}{\Delta t}$.
Step 2: Identify that the maximum acceleration occurs where the slope of the v-t graph is the steepest.
Step 3: From the given information for the steepest segment, the time interval is $\Delta t = 40 \text{ s} - 30 \text{ s} = 10 \text{ s}$.
Step 4: The corresponding change in velocity in this interval is $\Delta v = 80 \text{ cm/s} - 20 \text{ cm/s} = 60 \text{ cm/s}$.
Step 5: Substitute these values into the acceleration formula: $a_{\max} = \frac{60 \text{ cm/s}}{10 \text{ s}}$.
Step 6: Calculate the final value to get $a_{\max} = 6 \text{ cm s}^{-2}$.
Step 7: Conclude that the maximum acceleration of the moving object is $6 \text{ cm s}^{-2}$.

Step 1: Recall that the distance travelled by a particle is equal to the area under its velocity-time (v-t) graph.
Step 2: Divide the area under the v-t graph into four segments, each corresponding to a 1-second time interval.
Step 3: Calculate the area for the first interval (t=0 to t=1), which forms a triangle: Area = $\frac{1}{2} \times 1 \times 20 = 10 \text{ m}$.
Step 4: Calculate the area for the second interval (t=1 to t=2), which forms a rectangle: Area = $1 \times 20 = 20 \text{ m}$.
Step 5: Calculate the area for the third interval (t=2 to t=3), which forms a trapezium: Area = $\frac{1}{2} \times 1 \times (20 + 10) = 15 \text{ m}$.
Step 6: Calculate the area for the fourth interval (t=3 to t=4), which forms a rectangle: Area = $1 \times 10 = 10 \text{ m}$.
Step 7: Sum the areas of all four segments to find the total distance: Total Distance = $10 + 20 + 15 + 10 = 55 \text{ m}$.
Step 8: Conclude that the distance travelled by the particle in 4 seconds is 55 m.

Step 1: Recall that the distance travelled or height reached by an object is equal to the area under its velocity-time graph.
Step 2: The given velocity-time graph forms a trapezium.
Step 3: The area of a trapezium is given by the formula: Area = 1/2 * height * (sum of parallel sides).
Step 4: From the graph details, the maximum speed which represents the height of the trapezium is 3.6 m/s, and the parallel sides representing time intervals are 12 s and 8 s.
Step 5: Substitute the values into the formula: Height = 1/2 * 3.6 * (12 + 8).
Step 6: Calculate the result: Height = 1/2 * 3.6 * 20 = 36.0 m.
Step 7: Therefore, the height to which the lift takes the passengers is 36.0 m.

Step 1: Recall that the area under a velocity-time graph represents the change in position of the body.
Step 2: Displacement is defined as the net area, which is calculated as the area above the time axis minus the area below the time axis.
Step 3: Distance travelled is defined as the total path length, which is calculated as the sum of the absolute values of the areas above and below the time axis.
Step 4: Based on the given velocity-time graph, the net area evaluates to $8 \mathrm{m}$.
Step 5: The total absolute area under the graph evaluates to $16 \mathrm{m}$.
Step 6: Therefore, the displacement is $8 \mathrm{m}$ and the distance travelled is $16 \mathrm{m}$.
Step 7: This matches option A.

Step 1: The given displacement-time equation is $x = 2t + 1$.
Step 2: To find the velocity, differentiate the displacement $x$ with respect to time $t$: $v = \frac{dx}{dt} = \frac{d}{dt}(2t + 1)$.
Step 3: Evaluating the derivative gives $v = 2$, which is a constant value.
Step 4: A constant velocity means the $v-t$ graph is a horizontal straight line parallel to the time axis.
Step 5: Since the velocity is constantly $2$, the line intersects the velocity axis at $v = 2$, not at the origin $(0,0)$.
Step 6: Therefore, the corresponding $v-t$ graph is a straight line not passing through the origin.

Step 1: The particle is released from rest, which means its initial velocity is zero ($v = 0$ at $t = 0$). This implies the graph must pass through the origin.
Step 2: The particle falls freely under gravity, so it experiences a constant acceleration ($a = g$).
Step 3: Using the kinematic equation for velocity, $v = u + at$, and substituting $u = 0$ and $a = g$, we get $v = gt$.
Step 4: The equation $v = gt$ is a linear equation of the form $y = mx$, which represents a straight line passing through the origin.
Step 5: The slope of this line is $g$, which is a constant positive value, indicating that the velocity increases continuously with time.
Step 6: Therefore, the correct graphical representation is a straight line passing through the origin.

Step 1: When a particle is projected vertically upwards, it experiences a constant downward acceleration due to gravity ($a = -g$).
Step 2: The initial velocity is positive (directed upwards).
Step 3: The velocity decreases linearly with time according to the kinematic equation $v = u - gt$, which represents a straight line with a constant negative slope.
Step 4: At the maximum height, the velocity of the particle becomes zero.
Step 5: As the particle falls back down towards the ground, the velocity becomes negative (directed downwards) and its magnitude increases linearly with time.
Step 6: Therefore, the correct velocity-time graph is a straight line with a constant negative slope, starting from a positive value on the velocity axis, crossing the time axis, and continuing into the negative velocity region.

Step 1: Recall that the acceleration of a particle is given by the slope of its velocity-time (v-t) graph.
Step 2: The slope of the v-t graph is calculated using the formula $a = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}$.
Step 3: From the given description of the graph, the initial velocity $v_1 = 15 \text{ m/s}$ at $t_1 = 0 \text{ s}$ and the final velocity $v_2 = 0 \text{ m/s}$ at $t_2 = 3 \text{ s}$.
Step 4: Substitute these values into the slope formula: $a = \frac{0 - 15}{3 - 0}$.
Step 5: Calculate the acceleration: $a = \frac{-15}{3} = -5 \text{ ms}^{-2}$.
Step 6: Therefore, the acceleration of the particle is $-5 \text{ ms}^{-2}$.

Step 1: The average velocity of an object over a given time interval is defined as the total displacement divided by the total time taken, $v_{av} = \frac{\Delta x}{\Delta t}$.
Step 2: The displacement of the object is equal to the net area under the velocity-time (v-t) graph, which is the area above the time axis minus the area below the time axis.
Step 3: For the given v-t plot over the first 10 seconds, the magnitude of the area above the time axis is exactly equal to the magnitude of the area below the time axis.
Step 4: Because these two areas cancel each other out, the total displacement of the object during this 10-second interval is zero ($\Delta x = 0$).
Step 5: Substituting the zero displacement into the average velocity formula gives $v_{av} = \frac{0}{10} = 0$.
Step 6: Therefore, the average velocity of the object during the first 10 seconds is zero.

Step 1: Consider graph (I), a distance-time graph decreasing with time. Distance is the total path length covered, which can never decrease with time. Thus, this graph is not possible.
Step 2: Consider graph (II), a position-time graph with two values at a particular time. A particle cannot be at two different positions at the exact same instant. Thus, this graph is not possible.
Step 3: Consider graph (III), a velocity-time graph with two values at a particular time. A particle cannot have two different velocities at the exact same instant. Thus, this graph is not possible.
Step 4: Consider graph (IV), a speed-time graph showing negative speed. Speed is the magnitude of velocity and is a scalar quantity, which can never be negative. Thus, this graph is not possible.
Step 5: Since all the given graphs (I, II, III, and IV) violate the fundamental rules of motion, none of them can represent the one-dimensional motion of a particle.
Step 6: Therefore, all of the given graphs are invalid, making "All of these" the correct choice.

Step 1: Understand that velocity $v$ decreases linearly with displacement $s$, meaning the $v$-$s$ graph is a straight line.
Step 2: Identify the endpoints of this linear relationship. At $s = 0$, $v = 20 \text{ ms}^{-1}$ and at $s = 30 \text{ m}$, $v = 0 \text{ ms}^{-1}$.
Step 3: Calculate the slope of the $v$-$s$ graph, which corresponds to $\frac{dv}{ds}$.
Step 4: The slope is $\frac{dv}{ds} = \frac{\Delta v}{\Delta s} = \frac{0 - 20}{30 - 0} = -\frac{20}{30} = -\frac{2}{3} \text{ s}^{-1}$.
Step 5: Use the kinematic relation for acceleration in terms of velocity and displacement: $a = v \frac{dv}{ds}$.
Step 6: Substitute the given velocity $v = 10 \text{ ms}^{-1}$ and the calculated slope $\frac{dv}{ds} = -\frac{2}{3}$ into the formula.
Step 7: Calculate the acceleration: $a = 10 \times \left(-\frac{2}{3}\right) = -\frac{20}{3} \text{ ms}^{-2}$.
Step 8: Therefore, the acceleration of the particle at $v = 10 \text{ ms}^{-1}$ is $-\frac{20}{3} \text{ ms}^{-2}$.

Step 1: For uniformly accelerated motion, the relation between velocity and displacement is $v^2 = u^2 + 2as$.
Step 2: This equation is of the form $y = mx + c$, where $y = v^2$, $x = s$, slope $m = 2a$, and intercept $c = u^2$.
Step 3: Since the $v^2$ versus $s$ graph is a straight line, the acceleration is constant, confirming uniformly accelerated motion. So statement (A) is correct.
Step 4: The graph has a non-zero intercept on the $v^2$ axis, which means $u^2 \neq 0$, so the initial velocity $u \neq 0$. Therefore, statement (B) is wrong.
Step 5: For uniformly accelerated motion with non-zero initial velocity, the displacement-time relation is $s = ut + \frac{1}{2}at^2$, which is a parabolic equation. So statement (C) is correct.
Step 6: Since statement (B) is the only incorrect statement, the answer is (B).

From the kinematic equation of motion, we know that $v^2 = u^2 + 2as$.
This equation represents a straight line when $v^2$ is plotted against displacement $s$.
The slope of this $v^2$ versus $s$ graph is equal to $2a$.
From the given graph, the slope is calculated as $-\frac{16}{2} = -8 \mathrm{~ms}^{-2}$.
Therefore, we can equate the slope to $2a$, giving $2a = -8 \mathrm{~ms}^{-2}$.
Solving for acceleration $a$, we divide by 2 to get $a = -4 \mathrm{~ms}^{-2}$.
Hence, the acceleration of the particle is $-4 \mathrm{~ms}^{-2}$.

From the given acceleration-time graph, the acceleration is $a = 3 \mathrm{~ms}^{-2}$ for the time interval $t = 0$ to $t = 2 \mathrm{~s}$.
Since the particle starts from rest, the initial velocity $u = 0$.
Using the first equation of motion, the velocity at $t = 2 \mathrm{~s}$ is $v = u + at = 0 + (3)(2) = 6 \mathrm{~ms}^{-1}$.
For the next time interval from $t = 2 \mathrm{~s}$ to $t = 4 \mathrm{~s}$, the acceleration is $a = -3 \mathrm{~ms}^{-2}$.
The velocity at $t = 4 \mathrm{~s}$ is calculated as $v = 6 + (-3)(2) = 0 \mathrm{~ms}^{-1}$.
Therefore, the velocity-time graph consists of a straight line starting from the origin (0, 0) to the point (2, 6), followed by another straight line decreasing to the point (4, 0).
This matches the plot described in option A.