Motion In A Straight Line – NEET UG PhysicsPractice Questions & PYQs

Acceleration is fundamentally defined as the rate of change of velocity with respect to time, which, when expressed in terms of position, follows the relationship $a = v \frac{dv}{dx}$ where $v = \frac{dx}{dt}$. Starting with the relation $t=x^2+x$, we differentiate with respect to $x$ to get $\frac{dt}{dx} = 2x+1$, meaning velocity is the reciprocal $v = \frac{dx}{dt} = \frac{1}{2x+1}$, which can be written as $(2x+1)^{-1}$. Differentiating this expression with respect to $x$ using the chain rule yields $\frac{dv}{dx} = -1(2x+1)^{-2} \cdot 2 = -\frac{2}{(2x+1)^2}$. Substituting these results back into the acceleration formula gives $a = \left( \frac{1}{2x+1} \right) \cdot \left( -\frac{2}{(2x+1)^2} \right)$, simplifying to $-\frac{2}{(2x+1)^3}$.

To determine the kinematic properties of the particle, we use the fundamental relationships between position, velocity, and acceleration. Velocity is defined as the rate of change of position, which is obtained by differentiating the given equation $x = \alpha t^4 + \beta t^2 + \gamma t + \delta$ with respect to time, resulting in $v = 4 \alpha t^3 + 2 \beta t + \gamma$. Evaluating this expression at $t=0$ provides the initial velocity as $\gamma$.

Acceleration is the rate of change of velocity, which we find by differentiating the velocity function to obtain $a = 12 \alpha t^2 + 2 \beta$. At the initial time $t=0$, the acceleration simplifies to $2 \beta$. Combining these two results, the ratio of initial velocity to initial acceleration is $\frac{\gamma}{2 \beta}$.

The acceleration of an object is defined by the slope of its velocity-time graph, expressed mathematically as $a = \frac{dv}{dt}$. Initially, the motion starts from rest with a horizontal segment indicating zero slope, which means acceleration is zero, before transitioning into a phase of constant positive slope that results in a fixed positive acceleration. As the velocity levels off to a constant value, the acceleration returns to zero, followed by a final period where the downward slope of the graph creates a constant negative acceleration.

Velocity is the rate of change of position, determined by differentiating the position vector $\vec{r}(t) = 4t \hat{i} + 2t^2 \hat{j} + 5 \hat{k}$ to obtain the velocity vector $\vec{v}(t) = 4 \hat{i} + 4t \hat{j}$. At $t = 1$ s, this velocity vector becomes $\vec{v} = 4 \hat{i} + 4 \hat{j}$. The magnitude is calculated using $|\vec{v}| = \sqrt{4^2 + 4^2}$, which equals $4 \sqrt{2} \,\text{ms}^{-1}$. Because the velocity components in the x and y directions are identical, the angle $\theta$ relative to the x-axis satisfies $\tan \theta = 1$, resulting in a direction of $45^{\circ}$.

The motion of an object thrown vertically upward is governed by the kinematic relationship between displacement, initial velocity, time, and gravity. By assigning a positive sign to the upward initial velocity of 4 m/s and a negative sign to the acceleration due to gravity of 10 m/s^2, we can determine the final displacement relative to the starting point by applying the equation $s = ut + \frac{1}{2}at^2$. Substituting the values into this formula results in the expression $s = (4 \times 4) + (0.5 \times -10 \times 4^2)$, which simplifies to $s = 16 - 80$ and yields $s = -64$. Since the negative sign indicates the ball lands 64 meters below the point of release, the height of the bridge above the water is 64 meters.

The motion of the bullet inside the wooden block is governed by constant deceleration due to the uniform resistance offered by the material. This scenario allows us to apply the kinematic relationship $v^2 = u^2 + 2as$, where $v$ represents the final velocity, $u$ is the initial velocity, $a$ is the deceleration, and $s$ represents the distance covered. During the first part of the penetration, the velocity changes from $u$ to $u/3$ over a distance of 24 cm. Substituting these values into the kinematic equation gives $(\frac{u}{3})^2 = u^2 - 2a(24)$, which simplifies to an expression where $2a(24) = \frac{8u^2}{9}$.

For the remaining part of the block, the bullet starts with a velocity of $u/3$ and comes to a final velocity of 0 after traveling an additional distance $s$. Applying the same kinematic principle leads to $0 = (\frac{u}{3})^2 - 2as$, meaning $2as = \frac{u^2}{9}$. Comparing these two stages by taking the ratio of the two equations effectively cancels the acceleration and velocity terms, resulting in $\frac{2a(24)}{2as} = \frac{8u^2/9}{u^2/9}$. This simplifies to $\frac{24}{s} = 8$, revealing that the additional distance traveled is 3 cm. Combining the initial 24 cm with the additional 3 cm, the total length of the wooden block is 27 cm.

Average speed is defined as the total distance traveled divided by the total time taken, and for any journey split into two equal distance segments, this calculation simplifies to the harmonic mean of the speeds $v_1 = v$ and $v_2 = 2 v$. Because the vehicle spends more time traveling at the slower speed, the result is weighted toward the lower value rather than being a simple arithmetic mean. The calculation proceeds as follows:

$v_{\text{avg}} = \frac{2 \times v \times 2 v}{v + 2 v} = \frac{4 v}{3}$

Acceleration represents the rate of change of velocity, and a positive acceleration implies that $\frac{dv}{dt} > 0$, meaning the object's velocity is constantly increasing. Because the slope of a position-time graph corresponds to the instantaneous velocity, an increasing velocity dictates that the curve must become progressively steeper as time passes. This characteristic upward-bending shape is the geometric signature of positive acceleration.

The distance traveled by a freely falling object in any specific second $n$ is governed by the kinematic formula $S_n = u + \frac{1}{2}g(2n - 1)$. Given that the body starts from rest, the initial velocity $u$ is zero, which reduces the expression for distance to a proportion determined by the sequence of odd numbers. Applying this to each time interval, the distances for $n=1, 2, 3,$ and $4$ are calculated as $\frac{g}{2}(2(1) - 1)$, $\frac{g}{2}(2(2) - 1)$, $\frac{g}{2}(2(3) - 1)$, and $\frac{g}{2}(2(4) - 1)$, resulting in values of $\frac{g}{2}(1)$, $\frac{g}{2}(3)$, $\frac{g}{2}(5)$, and $\frac{g}{2}(7)$. Consequently, the common factor $\frac{g}{2}$ cancels out, revealing the ratio of distances covered in the first four seconds to be 1:3:5:7.

The velocity of a particle represented on a displacement-time graph is equivalent to the slope of the line, which is determined by the relationship $v = \tan \theta$, where $\theta$ is the angle the graph makes with the time axis. With this relationship, the velocity of the first particle is $\tan 30^{\circ}$ and the velocity of the second particle is $\tan 45^{\circ}$. The ratio of these two velocities is given by the expression $\frac{\tan 30^{\circ}}{\tan 45^{\circ}}$, which simplifies to $\frac{1/\sqrt{3}}{1}$, resulting in $1 : \sqrt{3}$.

Once an object is released from a moving vehicle, it inherits the vehicle's horizontal velocity at that instant and then acts as a projectile under the influence of gravity. When the ball is dropped at 4 s, its horizontal velocity is the same as the car's, which is $v = 0 + 5 \times 4 = 20$ m/s. During the 2 s interval between 4 s and 6 s, the ball continues with this constant horizontal velocity while its vertical velocity increases due to gravity according to $v_y = 10 \times 2 = 20$ m/s. The magnitude of the total velocity at 6 s is $v = \sqrt{20^2 + 20^2} = 20\sqrt{2}$ m/s. Since the ball is in free fall after being dropped, its acceleration is simply the constant downward acceleration due to gravity, which is 10 m/s².

The distance $S_n$ travelled by a body starting from rest under constant acceleration $a$ during the $n^{\text{th}}$ second is given by $S_n = \frac{a}{2}(2n-1)$. Since the block slides down a smooth incline, its acceleration remains constant at $a = g \sin \theta$, which makes the distance for the $n^{\text{th}}$ second $S_n = \frac{g \sin \theta}{2}(2n-1)$.

Evaluating the distance for the $(n+1)^{\text{th}}$ second requires substituting $n+1$ into the general formula, which yields $S_{n+1} = \frac{g \sin \theta}{2}(2(n+1)-1)$, simplifying to $S_{n+1} = \frac{g \sin \theta}{2}(2n+1)$. Dividing the expression for $S_n$ by the expression for $S_{n+1}$ cancels the common acceleration terms, leaving the ratio:

$\frac{S_{n}}{S_{n+1}} = \frac{2n-1}{2n+1}$

Since we are determining vertical distance under constant acceleration, we can rely on the kinematic relationship that connects velocity, acceleration, and height. The formula $v^2 = u^2 + 2gh$ effectively calculates the height, where $v$ is the final velocity, $u$ is the initial velocity, $g$ is the acceleration due to gravity, and $h$ is the vertical displacement.

To isolate the height, we rearrange the equation into $h = \frac{v^2 - u^2}{2g}$. Plugging \in the provided values of 80 m/s for final velocity, 20 m/s for initial velocity, and 10 m/s² for gravity, the expression becomes $h = \frac{80^2 - 20^2}{2 \times 10}$. Solving this calculation leads to $h = \frac{6400 - 400}{20}$, confirming the tower height is 300 meters.

Because the escalator and Preeti are moving in the same direction, their relative speeds combine, shortening the total time of travel. If we denote the length of the escalator as $L$, Preeti's walking speed is $v_1 = \frac{L}{t_1}$ and the escalator's speed is $v_2 = \frac{L}{t_2}$. By walking on the moving escalator, her effective speed becomes $v_{net} = \frac{L}{t_1} + \frac{L}{t_2}$, and the time taken to travel the distance $L$ is $t = \frac{L}{v_{net}}$. Substituting the speed values into this formula gives $t = \frac{L}{\frac{L}{t_1} + \frac{L}{t_2}}$, which simplifies to $\frac{t_1 t_2}{t_1 + t_2}$.

Calculating the distance travelled by an object requires integrating its velocity function over the specific time interval, as velocity is the time derivative of position. Given that $v = At + Bt^2$, you determine the total distance by evaluating the definite integral with limits from $1$ to $2$.

The integral is set up as follows:
$s = \int_{1}^{2} (At + Bt^2) dt$
Applying the power rule for integration leads to the following expression:
$s = \left[ \frac{At^2}{2} + \frac{Bt^3}{3} \right]_{1}^{2}$
Evaluating this expression by substituting the upper limit of $2$ and the lower limit of $1$ results in:
$s = \left( \frac{A(2)^2}{2} + \frac{B(2)^3}{3} \right) - \left( \frac{A(1)^2}{2} + \frac{B(1)^3}{3} \right)$
Simplifying these terms gives the final distance of $\frac{3A}{2} + \frac{7B}{3}$.

The instantaneous velocity of an object moving in a straight line is calculated by taking the first derivative of its position function with respect to time. For car P, differentiating $x_P(t) = at + bt^2$ with respect to time yields the velocity function $v_P(t) = a + 2bt$. Applying the same calculus principle to car Q, where the position is given by $x_Q(t) = ft - t^2$, results in the velocity function $v_Q(t) = f - 2t$.

Setting these two velocity functions equal identifies the specific time when the cars share the same velocity: $a + 2bt = f - 2t$. Rearranging the equation to collect all terms involving $t$ on one side gives $2bt + 2t = f - a$. Factoring $2t$ from the left side produces $2t(1 + b) = f - a$, which allows for the isolation of the time variable:
$t = \frac{f - a}{2(1 + b)}$

Acceleration in terms of position is determined by the relation $a = v\frac{dv}{dx}$, which accounts for the chain rule when velocity is defined as a function of position rather than time. Differentiating the velocity function $v = \beta x^{-2n}$ with respect to $x$ using the power rule yields the spatial derivative $\frac{dv}{dx} = -2n\beta x^{-2n-1}$. Substituting these components into the acceleration formula gives:
$a = (\beta x^{-2n})(-2n\beta x^{-2n-1})$
Multiplying these expressions, the constants combine to $-2n\beta^2$, and applying the laws of exponents by adding $-2n$ and $-2n-1$ results in the final acceleration:
$a = -2n\beta^2x^{-4n-1}$

For an object falling freely from rest, the distance covered in time $t$ is given by the kinematic equation $s = \frac{1}{2}gt^2$. Over the first 5 seconds, the distance covered is $h_1 = \frac{1}{2}g(5)^2 = \frac{25}{2}g$. The distance $h_2$ covered in the next 5-second interval is obtained by subtracting the distance covered in the first 5 seconds from the total distance covered in 10 seconds, which yields $h_2 = \frac{1}{2}g(10)^2 - \frac{25}{2}g = \frac{100}{2}g - \frac{25}{2}g = \frac{75}{2}g$, showing that $h_2 = 3h_1$. Similarly, the distance $h_3$ covered in the third 5-second interval is determined by subtracting the total distance covered in the first 10 seconds from the distance covered in 15 seconds, resulting in $h_3 = \frac{1}{2}g(15)^2 - \frac{100}{2}g = \frac{225}{2}g - \frac{100}{2}g = \frac{125}{2}g$, which is $5h_1$. Comparing these results, the distances $h_1$, $h_2$, and $h_3$ follow the ratio $1:3:5$, which simplifies to the relation $h_1 = \frac{h_2}{3} = \frac{h_3}{5}$.

The velocity of a particle is defined as the first derivative of its displacement with respect to time, meaning we must first express displacement as a function of time to identify the point where the velocity vanishes. By rearranging the given equation $t = \sqrt{x} + 3$ as $\sqrt{x} = t - 3$ and squaring both sides, we obtain the displacement formula $x = (t - 3)^2$. Differentiating this expression yields the velocity $v = \frac{dx}{dt} = 2(t - 3)$. Setting this velocity to zero indicates the particle is momentarily at rest when $t = 3$. Evaluating the displacement at this instant, $x = (3 - 3)^2$, reveals the particle's position is zero meters.

Position, velocity, and acceleration are fundamentally connected through calculus, where velocity is the derivative of position and acceleration is the derivative of velocity. For the position function $x = 8 + 12t - t^3$, the velocity is given by the derivative $v = 12 - 3t^2$. Setting $v = 0$ to determine the moment the particle stops reveals $12 - 3t^2 = 0$, which results in a time of 2 s. The acceleration is then found by differentiating the velocity to obtain $a = -6t$. Substituting 2 s into this expression yields an acceleration of -12 ms^-2, and because retardation corresponds to the magnitude of this value, the result is 12 ms^-2.