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

Average speed is defined as the total distance traveled divided by the total time taken for the entire journey. Because the object moves at different speeds over two equal distances, it spends a different amount of time traveling each half, which means we cannot simply calculate the arithmetic mean of the speeds. If we represent each half of the total distance as $x$, then the total distance is $2x$. The time taken to cover the first half is $t_1 = \frac{x}{v_1}$, and the time taken for the second half is $t_2 = \frac{x}{v_2}$. Combining these, the total time for the journey is $t_1 + t_2 = \frac{x}{v_1} + \frac{x}{v_2} = \frac{x(v_1 + v_2)}{v_1 v_2}$. Dividing the total distance $2x$ by this total time cancels out the distance variable $x$, resulting in the final expression for average speed as $\frac{2v_1 v_2}{v_1 + v_2}$.

When an object falls freely from rest, its final velocity increases based on the distance covered and the constant acceleration due to gravity. This motion is described by the kinematic equation:
$v^2 = u^2 + 2gh$
Since the stone starts from rest, its initial velocity $u$ is $0$. By substituting the given height of $20 \text{ m}$ and the acceleration due to gravity of $10 \text{ m s}^{-2}$ into the formula, we find that $v^2 = 0^2 + 2 \times 10 \times 20$. Solving this yields $v^2 = 400$, which indicates that the stone strikes the ground with a velocity of $20.0 \text{ m/s}$.

The kinematic equation $h = ut + \frac{1}{2}gt^2$ describes the distance fallen by an object under gravity, where $h$ represents the vertical distance, $u$ is the initial velocity, $t$ is the time elapsed, and $g$ is the acceleration due to gravity. Since both balls fall from the same platform and converge at the same point, they cover an identical vertical distance.

The first ball is dropped from rest and travels for 18 seconds. Its distance from the platform is calculated as:
$h = 0 \times 18 + \frac{1}{2} \times 10 \times 18^2 = 1620$
The second ball is released 6 seconds later, meaning it travels for 12 seconds by the time the meeting occurs. With an initial downward velocity $v$, the distance it covers is expressed as:
$h = v \times 12 + \frac{1}{2} \times 10 \times 12^2 = 12v + 720$
Because the total distance fallen is the same for both objects, we set these two expressions equal to each other:
$12v + 720 = 1620$
Subtracting 720 from both sides yields $12v = 900$, which simplifies to an initial velocity of 75 meters per second.

Kinematic definitions establish that acceleration is the second time derivative of position, whereas velocity is the first. Differentiating $x=(t+5)^{-1}$ with respect to time yields velocity $v=-(t+5)^{-2}$, and differentiating once more provides the acceleration $a=2(t+5)^{-3}$. To express acceleration in terms of velocity, we identify the relationship by raising the magnitude of velocity to the power of 3/2, which effectively isolates the time-dependent term $(t+5)^{-3}$. Substituting this into the acceleration formula reveals that the acceleration is directly proportional to $v^{3/2}$.

Because the particle is under a constant force, it moves with uniform acceleration from rest, where distance is proportional to the square of time according to the kinematic formula $s = \frac{1}{2}at^2$. Calculating the distance for the first 10 seconds gives $S_1 = \frac{1}{2} a (10)^2 = 50a$, while the distance for the first 20 seconds is $S_2 = \frac{1}{2} a (20)^2 = 200a$. Comparing these two values demonstrates that $200a$ is four times $50a$, establishing that $S_2 = 4S_1$.

The core concept is relative velocity. Because both vehicles are moving in the same direction, the scooterist effectively closes the gap by traveling at a speed equal to the difference between their individual velocities. The scooterist needs to cover an initial separation of 1000 m in 100 s, which means the necessary relative speed is the quotient of the distance and time, resulting in 10 m/s. Since this relative speed is defined as the difference between the scooter's velocity $v_s$ and the bus's velocity of 10 m/s, the relationship is described by the equation $v_s - 10 = 10$. Solving for the unknown variable, the scooterist must maintain a speed of 20 m/s.

Since the particle moves with constant acceleration, we can relate its change in velocity and displacement using the kinematic formula $v^2 = u^2 + 2as$. Substituting the initial velocity of 10 m/s, final velocity of 20 m/s, and displacement of 135 m into this formula yields $20^2 = 10^2 + 2 \cdot a \cdot 135$, which simplifies to $400 = 100 + 270a$, or $300 = 270a$, giving an acceleration of $a = 10/9$ m/s^2. With the acceleration determined, we can apply the velocity-time equation $v = u + at$ to solve for the time, where substituting the known values results in $20 = 10 + (10/9)t$. Simplifying this equation gives $10 = (10/9)t$, which results in a time of $t = 9$ s.

The distance travelled in the $n$th second is calculated by finding the difference between the total displacement after $n$ seconds and the total displacement after $n-1$ seconds. Given an initial velocity $u = 0$ and acceleration $a = 4/3$ m s$^{-2}$, the distance covered in three seconds is $\frac{1}{2} \cdot \frac{4}{3} \cdot 3^2 = 6$ meters, while the distance covered in two seconds is $\frac{1}{2} \cdot \frac{4}{3} \cdot 2^2 = 8/3$ meters. Subtracting the distance covered in two seconds from the distance covered in three seconds yields $6 - 8/3 = 10/3$ meters. Alternatively, using the formula $S_n = u + \frac{a}{2}(2n - 1)$ for the $n$th second directly provides the result as $0 + \frac{4/3}{2}(2 \cdot 3 - 1) = \frac{2}{3} \cdot 5 = 10/3$ meters.

Maximum speed occurs when the acceleration is zero, as the velocity reaches an extremum where its rate of change vanishes. By differentiating the position equation $x = 9t^2 - t^3$ with respect to time, we find the velocity $v = 18t - 3t^2$. Setting the derivative of velocity, $a = 18 - 6t$, equal to zero confirms that the particle reaches its maximum speed at $t = 3$ s. Substituting this time into the original position formula yields $x = 9(3)^2 - (3)^3$, which simplifies to 54 m.

Average speed is defined as the total distance traveled divided by the total time taken for the entire journey. If we represent the distance between points $X$ and $Y$ as $d$, the vehicle covers a total distance of $2d$ for the complete round trip. The time required for the outbound trip is $\frac{d}{v_u}$, while the time required for the return trip is $\frac{d}{v_d}$. Adding these two durations yields a total time of $\frac{d}{v_u} + \frac{d}{v_d}$, which can be simplified to $\frac{d(v_u + v_d)}{v_u v_d}$. By dividing the total distance of $2d$ by this total time, the expression becomes $\frac{2d}{\frac{d(v_u + v_d)}{v_u v_d}}$, which reduces to $\frac{2 v_u v_d}{v_u + v_d}$.

Velocity is the time integral of acceleration, representing the cumulative change in velocity over a given duration. Since the acceleration is governed by the function $f = f_0(1 - \frac{t}{T})$, the velocity is determined by integrating this expression over the interval starting from $t = 0$ until the acceleration ceases, which occurs at $t = T$.

The velocity $v_x$ is found by evaluating the definite integral:
$v_x = \int_0^T f_0 \left(1 - \frac{t}{T} \right) dt$
Calculating this integral yields $f_0 \left[ t - \frac{t^2}{2T} \right]_0^T$. Applying the limits of integration results in $f_0 (T - \frac{T^2}{2T})$, which simplifies to $f_0(T - \frac{1}{2}T)$, resulting in a final velocity of $\frac{1}{2} f_0 T$.

Average speed is determined by the total distance traveled over a given time interval, while average velocity is defined as the net displacement divided by that same time. For a car completing one full lap on a circular track with radius $r = 100 \text{ m}$, the total distance traveled is equal to the circumference of the circle, $2 \pi r$, while the net displacement is $0 \text{ m}$ because the car returns to its exact starting position.

To find the average speed, we use the values $r = 100 \text{ m}$ and $t = 62.8 \text{ s}$ to calculate the distance:
$\text{Average speed} = \frac{2 \times 3.14 \times 100}{62.8} = 10 \text{ m/s}$
Because the net displacement is $0 \text{ m}$, the average velocity is calculated as:
$\text{Average velocity} = \frac{0}{62.8} = 0 \text{ m/s}$
Thus, the average velocity and average speed for the lap are $0$ and $10 \text{ m/s}$ respectively.

When an object falls freely from a height under the influence of gravity, the time $t$ required to reach the ground is determined by the kinematic equation $t = \sqrt{\frac{2h}{g}}$, where $h$ is the height and $g$ is the acceleration due to gravity. Crucially, the mass of the objects does not affect the time of descent, so the $1\,\text{kg}$ and $3\,\text{kg}$ values are irrelevant to the outcome.

Because the gravitational acceleration $g$ is identical for both bodies, the ratio of the time taken for body A to body B simplifies to the square root of the ratio of their heights. By substituting the given heights of $16\,\text{m}$ and $25\,\text{m}$ into the relationship $\frac{t_A}{t_B} = \sqrt{\frac{h_A}{h_B}}$, we obtain $\sqrt{\frac{16}{25}}$, which simplifies to $\frac{4}{5}$.

The velocity of a particle moving along a straight line is defined as the time derivative of its position, and the particle comes to rest at the precise moment when this instantaneous velocity becomes zero. To determine the distance traveled, we differentiate the given displacement equation $x=40+12t-t^{3}$ with respect to time to obtain the velocity function $v = \frac{dx}{dt} = 12-3t^{2}$. Setting this expression to zero allows us to find the time at which motion ceases, yielding $12-3t^{2}=0$, which simplifies to $t^{2}=4$, or $t=2$ seconds.

We identify the particle's initial position by evaluating the displacement equation at $t=0$, which gives $x=40$ meters. We then calculate its position at the moment it comes to rest by substituting $t=2$ into the original equation, resulting in $x=40+12(2)-(2)^{3} = 40+24-8 = 56$ meters. The total distance traveled is the difference between these two positions, $56-40=16$ meters.

Velocity is defined as the time derivative of displacement, allowing us to find the particle's speed by differentiating $x=a e^{-\alpha t}+b e^{\beta t}$ to obtain $v = -a \alpha e^{-\alpha t} + b \beta e^{\beta t}$. As time moves forward, the first term $-a \alpha e^{-\alpha t}$ steadily approaches zero, while the second term $b \beta e^{\beta t}$ grows larger because both $b$ and $\beta$ are positive constants. This means the velocity is effectively dominated by the growing positive term, forcing the particle's speed to go on increasing as time progresses.

When an object is thrown vertically upward, its velocity decreases as it ascends until it stops momentarily at its peak. We can determine the maximum height by analyzing the segment of motion from the halfway point to the summit. During this interval, the final velocity is 0 m/s, the initial velocity is 10 m/s, the gravitational acceleration is 10 m/s squared, and the vertical distance covered is $h/2$. Applying the kinematic equation $v^{2} = u^{2} - 2gh$, we substitute these known values to get

$0 = 10^{2} - 2 \times 10 \times \frac{h}{2}$

Simplifying this equation yields $10h = 100$, which results in a maximum height of 10 m.

At the peak of the ball's trajectory, its velocity is zero. To find the distance covered in the final $t$ seconds of the ascent, one can view the motion in reverse by treating the peak as the starting point where the ball falls from rest under the influence of gravity. Applying the standard kinematic equation for distance covered from rest, the displacement is $h = \frac{1}{2}gt^2$. This physical symmetry confirms that the distance traversed against gravity in the final moments of rising is identical to the distance the ball would fall from the peak in the same duration.

To keep more than two balls airborne simultaneously, the first ball thrown must remain in the air for longer than the time required to release the third ball. Because balls are launched at intervals of 2 seconds, the first ball is thrown at time zero, the second at 2 seconds, and the third at 4 seconds. Consequently, the first ball must not have hit the ground before the third one is thrown, which means its total flight time must exceed 4 seconds.

Applying the formula for time of flight $T = \frac{2u}{g}$, the condition becomes $\frac{2u}{g} > 4$. Substituting the value $g = 9.8 \text{ m/s}^2$ into the inequality yields $\frac{2u}{9.8} > 4$. Solving this for $u$ results in $2u > 39.2$, which indicates the initial throwing speed must be greater than 19.6 m/s.

The motion of an object projected vertically is governed by the kinematic relation $v^2 = u^2 - 2gh$. At its maximum height $H$, the particle is at rest, meaning $0 = u^2 - 2gH$, which implies $u^2 = 2gH$. Given that the velocity is 10 m/s at half this height, $H/2$, we can apply the same kinematic equation to find that $10^2 = u^2 - 2g(H/2)$. Substituting $u^2 = 2gH$ into this equation simplifies to $100 = 2gH - gH$, or $100 = gH$. Using the given gravitational acceleration $g = 10$ m/s$^2$, the maximum height calculates to 10 m.

Understanding the motion of an object requires analyzing how its position changes over time using derivatives. Since velocity represents the rate of change of position and acceleration denotes the rate of change of velocity, we calculate acceleration by finding the second derivative of the position function $s$ with respect to time $t$.

First, differentiating $s = 3t^3 + 7t^2 + 14t + 8$ provides the velocity function:
$v = \frac{ds}{dt} = 9t^2 + 14t + 14$
Differentiating the velocity function again yields the expression for acceleration:
$a = \frac{dv}{dt} = 18t + 14$
Substituting $t = 1$ second into this expression results in $18(1) + 14$, which equals an acceleration of 32 m/s^2.