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

Acceleration is the rate of change of velocity, which is itself the rate of change of position with respect to time. By taking the first derivative of the position function $x = at^2 - bt^3$ with respect to time, the velocity is found to be $v = 2at - 3bt^2$. Differentiating this velocity expression provides the acceleration, which is determined to be $2a - 6bt$. Setting this value equal to zero results in the equation $2a - 6bt = 0$. Solving for $t$ gives $6bt = 2a$, which simplifies to $t = \frac{a}{3b}$.

For motion to occur in a single dimension, a particle must possess a unique velocity at any given instant of time $t$. Because a physical object cannot exist in two distinct states of motion simultaneously, any graph representing such movement must satisfy the condition where each input $t$ maps to exactly one value of velocity $v$. The curve presented in option B violates this principle by assigning multiple velocity values to a single instant, which is why it fails to represent one-dimensional motion.

Since acceleration is the time rate of change of velocity, we determine the instantaneous velocity by integrating the acceleration $a = bt$ with respect to time, which yields $v = \int bt \, dt = \frac{1}{2}bt^2 + C$. Applying the initial condition that velocity equals $v_{0}$ when $t = 0$ identifies the constant $C$ as $v_{0}$, resulting in the velocity function $v = v_{0} + \frac{1}{2}bt^2$. Because displacement is the integral of velocity with respect to time, integrating this expression gives $x = \int (v_{0} + \frac{1}{2}bt^2) dt = v_{0}t + \frac{1}{6}bt^3 + C'$. Given that the particle starts at the origin where $x = 0$ at $t = 0$, the constant $C'$ is zero, which means the distance traveled by the particle is $v_{0}t + \frac{1}{6}bt^3$.

For objects moving with uniform acceleration, the distance covered is equal to the product of the average velocity and the total time elapsed. Since the train increases its speed linearly from 20 km/h to 60 km/h, the average speed during this interval is determined by $\frac{20 + 60}{2} = 40$ km/h. By applying this average rate over the 4 hours of travel, the total distance covered is calculated as $40 \times 4 = 160$ km.

When a ruler is dropped from rest, the distance $s$ it covers relates to the time $t$ it takes to catch it according to the kinematic equation $s = \frac{1}{2}gt^2$. Because the acceleration due to gravity $g$ remains constant, a larger reaction time $t$ inevitably results in a greater distance $s$ fallen. By arranging the reaction times provided from largest to smallest-0.22 for B, 0.21 for E, 0.20 for A, 0.19 for D, and 0.18 for C-we determine the corresponding order of the distance traveled to be $B > E > A > D > C$.

The motion of the car is defined by two distinct phases: accelerating from rest to a peak velocity and then decelerating from that same peak velocity back to rest. Since the car reaches its maximum velocity $v$ at the precise moment acceleration ends and deceleration begins, this velocity serves as the common link between the two stages of the journey.

Defining the maximum velocity as $v$, the time taken to accelerate from rest at rate $\alpha$ is given by $t_1 = \frac{v}{\alpha}$. Similarly, the time taken to decelerate from $v$ to zero at rate $\beta$ is $t_2 = \frac{v}{\beta}$. Given that the total time elapsed is the sum of these intervals, we write $t = t_1 + t_2$, which simplifies to $t = \frac{v}{\alpha} + \frac{v}{\beta}$. Factoring out the velocity $v$ allows us to express the equation as $t = v \left( \frac{\alpha + \beta}{\alpha \beta} \right)$, and isolating $v$ provides the final value of $v = \frac{\alpha \beta t}{\alpha + \beta}$.

The distance an object covers during a specific second of its flight is determined by the equation $s_{n} = u + \frac{a}{2}(2n - 1)$. Because the body starts from rest, the initial velocity is $u = 0$, and the acceleration $a$ is equivalent to gravity $g$. Substituting these conditions into the formula for the 4th second where $n = 4$ yields $s_{4} = \frac{g}{2}(2 \times 4 - 1)$, which simplifies to $\frac{7g}{2}$. Applying the same approach for the 5th second by setting $n = 5$ results in $s_{5} = \frac{g}{2}(2 \times 5 - 1)$, leading to $\frac{9g}{2}$. Comparing these two outcomes establishes that the ratio of the distance moved is $7:9$.

Objects in free fall under gravity follow the motion described by the kinematic equation $h = \frac{1}{2} g t^2$, where the initial velocity is zero and the displacement $h$ depends on the square of the time $t$. To find the tower's height when a body falls a specific distance during its final interval, we define the distance covered in the last two seconds as the difference between the total height at time $t$ and the distance covered by the time the body reaches $t - 2$ seconds. Setting this difference equal to 40 meters leads to the equation $40 = \frac{1}{2} g t^2 - \frac{1}{2} g (t - 2)^2$.

Using $g = 10 \text{ m/s}^2$ in this equation allows us to simplify the expression for $t$. Substituting the value of $g$ yields $40 = 5 t^2 - 5 (t - 2)^2$. Expanding the term $(t - 2)^2$ results in $40 = 5 t^2 - 5 (t^2 - 4t + 4)$, which simplifies further to $40 = 5 t^2 - 5 t^2 + 20t - 20$, or $40 = 20t - 20$. Solving for $t$ gives $20t = 60$, meaning the total time of the fall is 3 seconds. Applying this time to the original height formula, $h = \frac{1}{2} \times 10 \times 3^2$, results in a total tower height of 45 meters.

Kinematic equations describe the motion of an object undergoing constant acceleration, where the displacement is determined by its initial velocity, final velocity, and the time elapsed. Since the final speed is provided in km/h, first convert it to standard units by multiplying by 5/18 to reach 40 meters per second. Given that the car starts from rest, the initial velocity is 0 meters per second, so we determine the acceleration using $a = \frac{v - u}{t}$, which simplifies to $a = 2$ meters per second squared. Applying these values to the distance formula $s = ut + \frac{1}{2}at^2$ gives $s = 0 + \frac{1}{2} \times 2 \times 20^2$, yielding a final distance of 400 meters.

Average speed is defined as the total distance covered divided by the total time taken for the entire trip. Since the car travels two equal distances at different speeds, the total time is the sum of the time spent on the first half and the time spent on the second half. With the total distance of 200 m covered at an average speed of 48 km/h, the relationship between these values is expressed by the equation $48 = \frac{200}{\frac{100}{40} + \frac{100}{v}}$. Dividing the total distance of 200 by the partial distances of 100 simplifies this expression to $48 = \frac{2}{\frac{1}{40} + \frac{1}{v}}$. Rearranging to solve for the unknown speed leads to $\frac{1}{40} + \frac{1}{v} = \frac{1}{24}$. Subtracting $\frac{1}{40}$ from $\frac{1}{24}$ yields $\frac{1}{v} = \frac{5}{120} - \frac{3}{120}$, which simplifies to the equation $\frac{1}{v} = \frac{1}{60}$. This indicates that the speed required for the second half of the journey is 60 km/h.

The motion of a particle is governed by the relationships between displacement, velocity, and acceleration, which are determined through calculus. Velocity is defined as the time derivative of displacement, and acceleration is the time derivative of velocity, meaning that finding the velocity at any specific condition requires calculating these derivatives sequentially.

Beginning with the displacement $s = t^{3}-6 t^{2}+3 t+4$, we differentiate with respect to time to obtain the velocity function $v = 3t^{2}-12t+3$. Differentiating again yields the acceleration function $a = 6t-12$. By setting this expression to zero, we solve $6t-12 = 0$ to find that the acceleration is zero at $t = 2$ seconds. Substituting this value back into the velocity equation gives $v = 3(2)^{2}-12(2)+3$, which simplifies to $v = -9$ m/s.

The instantaneous velocity of a particle on a displacement-time graph is defined by the slope of the curve at any specific point, calculated using the relationship $v = \frac{ds}{dt}$. A negative slope occurs when the displacement decreases as time progresses, signifying that the particle is moving in the opposite direction of the positive axis. Since the graph is sloping downwards specifically at point E, the instantaneous velocity at that location is negative.

When a ball is thrown vertically upward, the only force acting on it is gravity, which imposes a constant downward acceleration of $a = -g$ throughout the entire duration of the flight. This constant acceleration dictates that the velocity must change at a uniform rate, resulting in a velocity-time graph that takes the form of a straight line with a steady negative slope. Only plot C correctly depicts this continuous, unchanging negative gradient, which accurately reflects the physics of an object moving under the influence of gravity alone.

Average speed is defined as the total distance covered divided by the total time taken, so simply looking at the net displacement can lead to an error if the object reverses its direction. To find the true path length, first determine the velocity by differentiating the position function $s(t) = t^2 - 6t + 5$, which yields $v(t) = 2t - 6$. Since the velocity becomes zero at $t = 3$ s, the particle turns around at this instant, meaning we must compute the distance traveled in two separate intervals: from $t = 0$ s to $t = 3$ s and from $t = 3$ s to $t = 6$ s.

Evaluating the position function, the particle starts at 5 m, reaches -4 m at the turning point, and returns to 5 m by the end of the interval. The distance covered in the first segment is the magnitude of the change from 5 m to -4 m, which is 9 m, and the distance covered in the second segment is the magnitude of the change from -4 m to 5 m, which is also 9 m. Summing these distances gives a total path length of 18 m, which, when divided by the total time of 6 s, results in an average speed of 3 m/s.

Motion during a specific second is described by the equation
$S_n = u + \frac{a}{2}(2n - 1)$
where $n$ represents the specific time interval. Because the body begins from rest, we set the initial velocity $u$ to zero, simplifying the calculation for the fourth second to $S_4 = \frac{a}{2}(2 \times 4 - 1) = \frac{7a}{2}$. Similarly, substituting $n = 3$ for the third second yields $S_3 = \frac{a}{2}(2 \times 3 - 1) = \frac{5a}{2}$. Finding the ratio of these distances involves dividing $S_4$ by $S_3$, which eliminates the acceleration constant and results in $\frac{7}{5}$.

The stopping distance of a car is determined by the kinematic equation $v^2 - u^2 = 2as$, which indicates that since the final velocity is zero, the stopping distance $s$ is directly proportional to the square of the initial speed $u^2$. Given that 40 km/h is equivalent to $\frac{100}{9} \text{ m/s}$ and 80 km/h is $\frac{200}{9} \text{ m/s}$, doubling the speed results in the velocity term increasing by a factor of 2, and its square increasing by a factor of $2^2 = 4$. Consequently, the stopping distance must increase by a factor of 4 when the velocity doubles. Starting from an initial stopping distance of 2 m for the 40 km/h case, applying this factor results in a new minimum stopping distance of $2 \text{ m} \times 4$, which equals 8 m.

Average speed is determined by dividing the total distance traveled by the total time taken. When the journey is split into three equal portions, we can designate the total distance as $D$, making each segment equal to $\frac{D}{3}$. We calculate the time spent on each section by dividing the distance of that segment by the speed maintained during that part.

The time spent on the first segment is $\frac{D/3}{10} = \frac{D}{30}$, for the second part is $\frac{D/3}{20} = \frac{D}{60}$, and for the final portion is $\frac{D/3}{60} = \frac{D}{180}$. Summing these times results in a total time of $\frac{D}{30} + \frac{D}{60} + \frac{D}{180}$. Using a common denominator, this expression simplifies to $\frac{6D + 3D + D}{180}$, which reduces to $\frac{10D}{180}$, or simply $\frac{D}{18}$. Dividing the total distance $D$ by the total time $\frac{D}{18}$ leads to an average speed of 18 km/h.

Average speed is defined as the total distance covered divided by the total time taken, which requires determining the duration spent on each segment of the journey. If we represent the total distance as $D$, the car covers $D/2$ at 40 km/h and the remaining $D/2$ at 60 km/h. This means the time for the first part is $D/80$ and the time for the second part is $D/120$, resulting in a total time of $D/80 + D/120 = 5D/240 = D/48$. Dividing the total distance $D$ by the total time $D/48$ results in an average speed of 48 km/h.

The motion of the water drops is governed by the kinematic formula for free fall $h = \frac{1}{2}gt^2$, where $g$ represents the acceleration due to gravity, which we approximate as $10$ m/s$^2$. Given a total height of 5 m, the time taken for the first drop to reach the ground is calculated by solving $5 = \frac{1}{2} \times 10 \times t^2$, which yields $t = 1$ s. Because the third drop departs exactly when the first hits the ground, the 1 s interval covers the departure of three drops, meaning the time difference between consecutive drops is 0.5 s.

At the moment of impact for the first drop, the second drop has been falling for only 0.5 s. The distance it has traveled from the tap is calculated as $d = \frac{1}{2} \times 10 \times 0.5^2$, resulting in 1.25 m. Subtracting this fallen distance from the original height of 5 m confirms that the second drop is 3.75 m above the ground.

The distance between two consecutive buses remains constant and is defined by the product of the bus speed and the regular service interval, expressed as $d = v T$. This spatial separation persists regardless of whether an observer travels alongside or against the buses, though the relative speeds experienced differ based on the direction of travel.

In the direction of motion, the scooty moves with the bus, making the relative speed $(v - 60) \text{ km/h}$. Since the buses pass every 30 minutes, or $0.5 \text{ hours}$, the gap is $0.5(v - 60)$. When moving against oncoming traffic, the relative speed is $(v + 60) \text{ km/h}$, and with a 10-minute interval, or $\frac{1}{6} \text{ hours}$, the gap is $\frac{1}{6}(v + 60)$. Equating these two expressions for the distance, we have:
$0.5(v - 60) = \frac{1}{6}(v + 60)$
Multiplying both sides by 6 simplifies this to $3(v - 60) = v + 60$, which results in a bus speed of $v = 120 \text{ km/h}$. Substituting this speed back into the distance calculation gives $d = (120 - 60) \times 0.5 = 30 \text{ km}$. Applying the relation $d = v T$, we solve $30 = 120 \times T$ to find that $T = 0.25 \text{ hours}$, which is 15 minutes.