Motion In A Straight LineClass 11 Physics NCERT Solutions

An object can be considered a point object if its size is much smaller than the distance it travels in a reasonable duration of time.

(a) Point object: The size of a railway carriage is very small compared to the distance between two stations. Therefore, the carriage can be treated as a point object.

(b) Point object: The size of the monkey is negligible compared to the circumference of the circular track. Therefore, the monkey can be considered a point object.

(c) Not a point object: The size of the spinning cricket ball is comparable to the distance over which it turns on hitting the ground. The spin and the turning motion are related to its size, so it cannot be considered a point object.

(d) Not a point object: The size of the beaker is comparable to the height of the table from which it slips. The tumbling motion involves different parts of the beaker covering different distances, so it cannot be treated as a point object.

(a) (A) lives closer to the school than (B): From the graph, the position of home P for child A is $OP$, and the position of home Q for child B is $OQ$. Since $OP < OQ$, A lives closer to the school.

(b) (A) starts from the school earlier than (B): A starts from the school (origin O) at time $t=0$, while B starts from the school at a later time (when $t > 0$). Therefore, A starts earlier.

(c) (B) walks faster than (A): The speed of the children can be determined by the slope of their respective $x-t$ graphs. Since the slope of the graph for B is steeper than the slope of the graph for A, B walks faster than A.

(d) A and B reach home at the (different) time: The graph shows that A reaches home P at time $t_A$ and B reaches home Q at time $t_B$. Clearly, $t_A < t_B$. So, they reach home at different times.

(e) (B) overtakes (A) on the road (once): B starts later but walks faster. The point where the two graphs intersect indicates the point where they are at the same position at the same time. Since the graphs intersect once, B overtakes A on the road once.

Plotting the x-t graph: We can represent time on the x-axis and position on the y-axis.

The resulting graph will show a line with a positive slope, followed by a horizontal line, and then a line with a steep negative slope returning to the time axis.

Calculation of time to fall in the pit: Let's track the drunkard's position over time.

At the end of 4 cycles, the drunkard is at the 8 m mark after 32 s. From this point, he takes 5 steps forward. To cover the remaining distance to the pit (13 m - 8 m = 5 m), he needs to take 5 forward steps.

Total time taken = Time for 4 cycles + Time for the final 5 forward steps Total time = 32 s + 5 s = 37 s.

Final Answer: The drunkard takes 37 s to fall into the pit.

Step 1: Convert initial speed to m/s $v_0 = 126 \text{ km h}^{-1} = 126 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 35 \text{ m s}^{-1}$

Step 2: Calculate the retardation (a) Formula: $v^2 = v_0^2 + 2ax$

Calculation: $0^2 = (35)^2 + 2 \times a \times 200$ $0 = 1225 + 400a$ $400a = -1225$ $a = -\frac{1225}{400} = -3.0625 \text{ m s}^{-2}$ The retardation is the magnitude of this acceleration, which is $3.0625 \text{ m s}^{-2}$.

Step 3: Calculate the time taken (t) Formula: $v = v_0 + at$

Calculation: $0 = 35 + (-3.0625)t$ $3.0625t = 35$ $t = \frac{35}{3.0625} \approx 11.43 \text{ s}$

Final Answer: The retardation of the car is $3.06 \text{ m s}^{-2}$, and it takes approximately $11.43 \text{ s}$ for the car to stop.

(a) The acceleration during the upward motion is the acceleration due to gravity, which is always directed vertically downwards.

(b) At the highest point of its motion, the ball is momentarily at rest, so its velocity is $0 \text{ m s}^{-1}$. The acceleration is still the acceleration due to gravity, which is $9.8 \text{ m s}^{-2}$ directed vertically downwards.

(c) According to the given convention:

(d) Let's use the standard convention where the point of projection is the origin and the upward direction is positive. Given:

To find the maximum height (h): Formula: $v^2 = v_0^2 + 2ah$ Calculation: $0^2 = (29.4)^2 + 2(-9.8)h$ $0 = 864.36 - 19.6h$ $19.6h = 864.36$ $h = \frac{864.36}{19.6} = 44.1 \text{ m}$

To find the time to return to the player's hands: First, find the time to reach the maximum height ($t_{up}$). Formula: $v = v_0 + at$ Calculation: $0 = 29.4 + (-9.8)t_{up}$ $9.8t_{up} = 29.4$ $t_{up} = \frac{29.4}{9.8} = 3 \text{ s}$ The time taken to fall back to the initial position is equal to the time taken to rise. So, time of descent $t_{down} = 3 \text{ s}$. Total time of flight = $t_{up} + t_{down} = 3 \text{ s} + 3 \text{ s} = 6 \text{ s}$.

Final Answer: The ball rises to a height of 44.1 m, and it returns to the player's hands after 6 s.

(a) True. A particle can have zero speed and non-zero acceleration. For example, when a ball is thrown vertically upwards, its speed becomes zero at the highest point of its trajectory, but it still has an acceleration equal to the acceleration due to gravity ($g = 9.8 \text{ m s}^{-2}$) acting downwards.

(b) False. Speed is the magnitude of velocity. If the speed is zero, the magnitude of velocity is zero, which means the velocity itself must be zero. It is not possible for a particle to have zero speed and non-zero velocity.

(c) True. In one-dimensional motion, constant speed means the magnitude of velocity is constant. Since the motion is in a straight line, if the particle does not reverse its direction, its velocity is constant. Constant velocity implies that the acceleration (rate of change of velocity) is zero. If the particle reverses direction, its speed is not constant at the point of reversal.

(d) False. A positive value of acceleration does not necessarily mean the particle is speeding up. Speeding up or slowing down depends on the relative signs of velocity and acceleration. If acceleration is positive and velocity is also positive, the particle speeds up. However, if acceleration is positive and velocity is negative (i.e., they are in opposite directions), the particle slows down. For example, if the positive direction is to the right, an object moving to the left (negative velocity) with a positive acceleration (directed to the right) will slow down.

Let's analyze the motion of the ball in parts. We take the downward direction as positive. $g = 9.8 \text{ m s}^{-2}$.

The graph consists of a series of straight lines forming triangular shapes, with the peak speed decreasing after each bounce.

(a) Magnitude of Displacement vs. Total Path Length (Distance)

Example: A person walks 40 m east and then turns around and walks 10 m west.

(b) Magnitude of Average Velocity vs. Average Speed

Example: Using the same example as above, let the person take 10 seconds for the entire journey.

Relationship and Condition for Equality The total path length is always greater than or equal to the magnitude of the displacement. $\text{Total Path Length} \geq |\text{Displacement}|$ Dividing both sides by the time interval ($\Delta t$), we get: $\frac{\text{Total Path Length}}{\Delta t} \geq \frac{|\text{Displacement}|}{\Delta t} = |\frac{\text{Displacement}}{\Delta t}|$ $\text{Average Speed} \geq |\text{Average Velocity}|$ The equality sign holds true only when the particle moves along a straight line in a single direction without any reversal. In this case, the total path length is equal to the magnitude of the displacement.

The instantaneous velocity is defined as the limit of the average velocity as the time interval $\Delta t$ approaches zero. $v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}$ Similarly, the instantaneous speed is the limit of the average speed as the time interval $\Delta t$ approaches zero. $\text{Instantaneous Speed} = \lim_{\Delta t \to 0} \frac{\text{Path Length}}{\Delta t}$ When the time interval $\Delta t$ is infinitesimally small, the particle does not have time to change its direction of motion. In this extremely small interval, the path length covered by the particle becomes equal to the magnitude of its displacement, $|\Delta x|$.

Therefore, $\lim_{\Delta t \to 0} \frac{\text{Path Length}}{\Delta t} = \lim_{\Delta t \to 0} \frac{|\Delta x|}{\Delta t} = |\lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}| = |v|$ This means that the instantaneous speed is always equal to the magnitude of the instantaneous velocity. The distinction between distance and displacement magnitude vanishes for an infinitesimally small time interval.

(a) Cannot represent one-dimensional motion. The x-t graph shows that for some values of time, the particle has two different positions. This is physically impossible, as a particle cannot be at two places at the same time.

(b) Cannot represent one-dimensional motion. The v-t graph shows that for some values of time, the particle has two different velocities (one positive and one negative). This is physically impossible.

(c) Cannot represent one-dimensional motion. The speed-time graph shows negative values for speed. Speed is the magnitude of velocity and can never be negative.

(d) Cannot represent one-dimensional motion. The graph shows that the total path length decreases with time for some intervals. The total path length covered by a particle can never decrease; it can either increase or remain constant.

No, it is not correct to say that the particle moves on a parabolic path for $t > 0$. The graph is an $x-t$ plot, which describes the position of the particle along a single axis (one dimension) as a function of time. It does not represent the trajectory or path of the particle in space.

The particle moves only along a straight line (one-dimensional motion) for all time, both $t < 0$ and $t > 0$.

Suitable physical context: A simple example could be an object, like a toy car, moving with a constant velocity on a frictionless surface. At $t=0$, it begins to accelerate uniformly, perhaps by a fan switching on or by moving onto a tilted surface. The entire motion occurs along a single straight line.

Step 1: Convert all speeds to m/s $v_P = 30 \times \frac{1000}{3600} = \frac{25}{3} \text{ m s}^{-1} \approx 8.33 \text{ m s}^{-1}$ $v_T = 192 \times \frac{1000}{3600} = \frac{160}{3} \text{ m s}^{-1} \approx 53.33 \text{ m s}^{-1}$

Step 2: Calculate the absolute speed of the bullet The absolute speed of the bullet ($v_B$) with respect to the ground is the speed of the van plus the muzzle speed of the bullet, as both are in the same direction. $v_B = v_P + v_M = \frac{25}{3} + 150 = \frac{25 + 450}{3} = \frac{475}{3} \text{ m s}^{-1} \approx 158.33 \text{ m s}^{-1}$

Step 3: Calculate the relative speed of the bullet with respect to the thief's car The speed relevant for damaging the thief's car is the speed with which the bullet hits the car, which is the relative speed of the bullet with respect to the thief's car ($v_{BT}$). Since both are moving in the same direction, we subtract their speeds. Formula: $v_{BT} = v_B - v_T$

Calculation: $v_{BT} = \frac{475}{3} - \frac{160}{3} = \frac{315}{3} = 105 \text{ m s}^{-1}$

Final Answer: The bullet hits the thief's car with a speed of $105 \text{ m s}^{-1}$.

(a) x-t graph: This graph shows the position of an object changing over time. The position increases, then decreases to zero, then increases again to a smaller value, and so on. This can represent a ball dropped from a certain height onto a hard floor. The ball hits the floor (x=0), bounces back up to a lesser height, falls again, and continues to bounce with decreasing height until it comes to rest.

(b) v-t graph: This graph shows velocity changing abruptly from positive to negative at regular intervals, while decreasing linearly in between. This is an idealized situation. A plausible, though not perfectly matching, scenario could be a ball bouncing between two walls. However, the constant change in velocity magnitude is hard to justify. A more abstract example is an object whose velocity is repeatedly and instantly reversed and slightly reduced, which is not common in simple mechanics. A better physical situation could be a ball thrown vertically upwards, which hits the ceiling, reverses direction, hits the floor, reverses direction again, and so on, assuming each collision is perfectly elastic but some energy is lost over time, which contradicts the constant peak velocity magnitude. Correction: A better interpretation for (b) is a ball thrown vertically upwards ($v$ decreases linearly to 0), falls back down ($v$ becomes negative and increases in magnitude), and is then hit by a bat at the initial position to send it back up with the same initial velocity. The instantaneous changes in velocity represent the impact with the bat.

(c) a-t graph: This graph shows acceleration being zero most of the time, with very short, large positive spikes at regular intervals. This represents an object moving with constant velocity (zero acceleration) that receives a very brief, strong push or kick in the direction of motion at regular intervals. For example, a ball being hit repeatedly by a bat, or a hammer striking a nail in short bursts.

In an $x-t$ graph: