Class 11 Physics (Physics Part-I) Chapter 2 Motion In a Straight Line

Key Concept: Instantaneous vs Average

c) It is the slope of the tangent line

[Solution Description] Instantaneous velocity is best described as the speed or velocity of an object at a specific moment in time, which corresponds to the slope of the tangent at that point on the position-time graph.

Key Concept: Real-World Application

a) 37 m/s

[Solution Description] Differentiate the position function to find velocity:

$v(t) = \frac{dx}{dt} = \frac{d}{dt}(4t^2 – 3t + 5) = 8t – 3.$

Substitute $t = 5$:

$v(5) = 8 \times 5 – 3 = 40 – 3 = 37 \text{ m/s}.$

Key Concept: Initial Velocity Calculation

d) 24 m/s

[Solution Description]

Using the equation $x = v_0 t + \frac{1}{2} a t^2$ and the relation $v = v_0 + at$, we know:

$x = 90 \text{ m},\; t = 6 \text{ s},\; v = 30 \text{ m/s}$

Solving $v = v_0 + at$ for $a$:

$a = \frac{v – v_0}{t}$

Substitute into $x = v_0 t + \frac{1}{2} a t^2$:

$90 = v_0(6) + \frac{1}{2} \left(\frac{30 – v_0}{6}\right)(36)$

Simplifying:

$90 = 6v_0 + 18 – 3v_0$

$90 = 3v_0 + 18$

$72 = 3v_0$

$v_0 = 24 \text{ m/s}$

So, the initial velocity was 24 m/s.

Key Concept: Basic Definitions

b) The total path length traveled by an object

[Solution Description] Distance is defined as a scalar quantity that represents the total path length traveled by an object, regardless of direction.

Key Concept: Constant Acceleration

c) A straight line inclined to the time axis

[Solution Description] For an object with constant acceleration, the velocity-time graph is a straight line inclined to the time axis. This is because the velocity changes uniformly over time.

Key Concept: Area Under Curve

c) 60 m

[Solution Description]

The area under a velocity-time graph gives the displacement of the object. As the area between $t=0$ and $t=10$ seconds is $60 \, \text{m}$, the displacement is $60 \, \text{m}$.

Key Concept: Basic Definition

c) 20 km

[Solution Description] The path length is the total distance traveled irrespective of direction. Here, the car first goes 8 km north and then 12 km south. So, the total path length is $8 \, \text{km} + 12 \, \text{km} = 20 \, \text{km}$.

Key Concept: Changing Frames

b) 57 km/h, west

[Solution Description]

To determine the passenger’s velocity relative to an observer on the road, we need to subtract the passenger’s velocity from the bus’s velocity since they are moving in opposite directions.

Let the positive direction be westward. Then, the bus’s velocity ($v_b$) is $$+60$$ km/h, and the passenger’s velocity ($v_p$) relative to the bus is $-3$ km/h.

Thus, the passenger’s velocity $v_r$ relative to the observer is:

$v_r = v_b + v_p = 60 – 3 = 57 \text{ km/h (westward)}$

The passenger’s velocity relative to an observer on the road is therefore 57 km/h to the west.

Key Concept: Complex Derivative Problem, Calculus-Based Problem

c) 26 $\text{m/s}^2$

[Solution Description]

To find the instantaneous acceleration, we first need to determine the velocity function by differentiating the position function with respect to time $t$. The position function is given as:

$x(t) = 3t^3 – 5t^2 + 7$

Differentiating $x(t)$ with respect to $t$, we get the velocity function $v(t)$:

$v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t^3 – 5t^2 + 7) = 9t^2 – 10t$

Next, we differentiate the velocity function to find the acceleration function $a(t)$:

$a(t) = \frac{dv}{dt} = \frac{d}{dt}(9t^2 – 10t) = 18t – 10$

Now, substituting $$t = 2$$ s into the acceleration function:

$a(2) = 18(2) – 10 = 36 – 10 = 26 \ \text{m/s}^2$

Therefore, the instantaneous acceleration at $t = 2$ s is $26 \ \text{m/s}^2$.

Key Concept: Basic Concept

b) 9.8 m/s²

[Solution Description] The acceleration due to gravity, $$g$$, is a constant value that represents the rate of acceleration for any object in free fall near the Earth’s surface. This value is approximately $$9.8 \, \text{m/s}^2$$.

Key Concept: Complex Problem Solving, Integration of Concepts

c) 29.5 m/s

[Solution Description] To find the average velocity, first determine the displacement $\Delta x = x(4) – x(2)$:

$x(4) = 4(4)^3 – 12(4)^2 + 9(4) + 1 = 64 \text{ m}$

$x(2) = 4(2)^3 – 12(2)^2 + 9(2) + 1 = 5 \text{ m}$

Then, $\Delta x = 64 – 5 = 59$ m. The time interval $\Delta t = 4 – 2 = 2$ s.

Now calculate the average velocity:

$v_{avg} = \frac{\Delta x}{\Delta t} = \frac{59}{2} = 29.5 \text{ m/s}$

Therefore, the average velocity is 29.5 m/s.

Key Concept: Formula Identification

a) $$x = v_0t + \frac{1}{2}at^2$$

[Solution Description] For uniformly accelerated motion, one of the kinematic equations for displacement is given by $$x = v_0t + \frac{1}{2}at^2$$.

Key Concept: Real-World Speed Application, Speed and Distance Ratio

c) 10 km

[Solution Description] The train starts from rest, so its initial speed $u = 0$. Final speed $v = 120$ km/h = $\dfrac{120 \times 1000}{60 \times 60}$ m/s = $33.33$ m/s. Time $t = 10$ minutes = $600$ seconds. The acceleration $a$ is given by $a = \dfrac{v – u}{t} = \dfrac{33.33 – 0}{600} = 0.0556$ m/s\textsuperscript{2}. Using the equation for distance $s$, $s = ut + \dfrac{1}{2}at^2 = 0 + \dfrac{1}{2} \times 0.0556 \times (600)^2 = 10000$ meters or $10$ km.

Key Concept: Equation Derivation

a) $9t^2 – 12t$

[Solution Description] To find the instantaneous acceleration, we need to take the derivative of the velocity function with respect to time. The given velocity function is $v = 3t^3 – 6t^2 + 4$.

Differentiating with respect to $t$, we get:

$a = \frac{dv}{dt} = \frac{d}{dt}(3t^3 – 6t^2 + 4) = 9t^2 – 12t.$

So, the expression for instantaneous acceleration is $a = 9t^2 – 12t$.

Key Concept: Concept Choice

c) The choice of origin and positive direction is arbitrary.

[Solution Description] The choice of origin and positive direction is arbitrary, meaning it can be chosen freely according to convenience. This flexibility allows for different perspectives or reference frames depending on the problem’s requirement.

Key Concept: Area Under Curve

b) Displacement

[Solution Description] The area under a velocity-time graph represents the displacement of the object. This is because integrating velocity with respect to time gives displacement.

Key Concept: Complex Calculation, Derivative Application

c) 28 m/s

[Solution Description]

First, we find the position at $t = 1$ s:

$x(1) = 4(1)^3 – 6(1)^2 + 2 = 0$

Next, we find the position at $t = 3$ s:

$x(3) = 4(3)^3 – 6(3)^2 + 2 = 56$

The displacement $\Delta x$ is $x(3) – x(1) = 56 – 0 = 56$$ m.

The time interval $\Delta t$ is $3 – 1 = 2$ s.

Therefore, the average velocity $v_{avg}$ is given by:

$v_{avg} = \frac{\Delta x}{\Delta t} = \frac{56}{2} = 28 \text{ m/s}$

Key Concept: Relative Motion with Acceleration, Multi-step Calculations

b) 16.67 seconds

[Solution Description] To find when the velocities are equal, we convert the first train’s velocity into m/s: $60 \text{ km/h} = \frac{60 \times 1000}{3600} \text{ m/s} = 16.67 \text{ m/s}$.

The velocity of the second train after time t seconds is given by v = u + at, where u = 0, $$a = 1 \text{ m/s}^2$$. So, v = t. We equate this with the first train’s velocity:

t = 16.67

Therefore, t = 16.67 seconds.

Key Concept: Average Velocity

b) $0 \, \text{m/s}$

[Solution Description] Average velocity is calculated using $\frac{x(t_2) – x(t_1)}{t_2 – t_1}$. We evaluate $x(t)$ at $t_1 = 1 \, \text{s}$ and $t_2 = 3 \, \text{s}$:

$x(1) = 4(1) – (1)^2 = 4 – 1 = 3 \, \text{m}$,

$x(3) = 4(3) – (3)^2 = 12 – 9 = 3 \, \text{m}$.

Average velocity:

$v_{avg} = \frac{3 \, \text{m} – 3 \, \text{m}}{3 \, \text{s} – 1 \, \text{s}}$

$= \frac{0 \, \text{m}}{2 \, \text{s}}$

$= 0 \, \text{m/s}$.

The average velocity is $0 \, \text{m/s}$.

Key Concept: Unit Recognition

d) Newton

[Solution Description]

The unit of force in the International System of Units (SI) is the Newton (N). This is derived from the equation F = ma, where F is force, m is mass in kilograms (kg), and a is acceleration in meters per second squared (m/s$^2$).

Key Concept: Basic Formula

d) All of the above

[Solution Description]

Kinematic equations for uniformly accelerated motion include the formulas that relate initial velocity ($v_0$), final velocity ($v$), acceleration ($a$), time ($t$), and displacement ($x$). The correct options are:

– $v = v_0 + at$

– $x = v_0t + \frac{1}{2}at^2$

– $v^2 = v_0^2 + 2ax$

Each one describes a different aspect of motion under constant acceleration conditions.

Key Concept: Non-Uniform Acceleration Analysis

b) 2 m/s²

[Solution Description]

The average acceleration can be calculated using the formula for average acceleration $a = \Delta v / \Delta t$. Here, $\Delta v = 25 – 5 = 20$ m/s, and $\Delta t = 10$ s. Thus, the average acceleration is $a = 20 / 10 = 2$ m/s².

Key Concept: Kinematic Equation

c) $x = v_0t + \frac{1}{2}at^2$

[Solution Description]

The kinematic equation representing displacement for uniformly accelerated motion is given by $x = v_0t + \frac{1}{2}at^2$, where x is the displacement, $v_0$ is the initial velocity, $$a$$ is the acceleration, and t is the time.

Key Concept: Calculating Velocity

c) 18 m/s

[Solution Description]

The position function is given as $x = 3t^2 + 5$. To find the velocity, we need to differentiate this position function with respect to time t

$v = \frac{dx}{dt} = \frac{d}{dt}(3t^2 + 5)$

Differentiating term by term:

v = 6t

Now substituting t = 3 seconds into this equation, we get:

$v = 6(3) = 18 \text{ m/s}$

Thus, the velocity at t = 3 s is 18 m/s.

Key Concept: Instantaneous vs Average

c) Instantaneous velocity equals average velocity

[Solution Description] When a car starts from rest and accelerates uniformly, its velocity increases linearly over time. At the midpoint of the time interval, which is 5 seconds in this case, the instantaneous velocity will be exactly half of the final velocity due to uniform acceleration.

Since the car reaches 20 m/s at 10 seconds, the instantaneous velocity at 5 seconds would be:

$v_{\text{instant}} = \frac{1}{2} \times 20 = 10 \text{ m/s}$

The average velocity for the entire duration from start to end is calculated as:

$v_{\text{average}} = \frac{\text{total distance}}{\text{total time}} = \frac{0 + 20}{2} = 10 \text{ m/s}$

In this particular case, both the instantaneous velocity at 5 seconds and the average velocity over the 10-second period are equal at 10 m/s.

Key Concept: Definition Check

d) Motion along a straight line

[Solution Description] Rectilinear motion refers to the motion of an object along a straight line. The term “rectilinear” indicates linear or straight path.

Key Concept: Units Identification

b) m/s$^2$

[Solution Description]

Acceleration is defined as the change in velocity per unit time. In SI units, velocity is measured in meters per second (m/s) and time is measured in seconds (s). Therefore, the SI unit of acceleration is meters per second squared (m/s$^2$).

Key Concept: Application of Formula

b) 2 m/s$^2$

[Solution Description] The average acceleration $a$ can be calculated using the formula:

$a = \frac{\Delta v}{\Delta t}$

where $\Delta v = v_f – v_i = 50 \, \text{m/s} – 20 \, \text{m/s} = 30 \, \text{m/s}$, and $\Delta t = 15 \, \text{s}$.

Substituting these values into the formula gives:

$a = \frac{30 \, \text{m/s}}{15 \, \text{s}} = 2 \, \text{m/s}^2$

Hence, the average acceleration is $2 \, \text{m/s}^2$.

Key Concept: Resultant Velocity

c) 15.81 km/h

[Solution Description]

The resultant velocity $V_r$ of the boat can be calculated using the Pythagorean theorem since the velocities form a right triangle:

$V_r = \sqrt{V_b^2 + V_c^2}$

where $V_b = 15 \, \text{km/h}$ is the velocity of the boat in still water, and $V_c = 5 \, \text{km/h}$ is the velocity of the current.

$V_r = \sqrt{15^2 + 5^2}$

$V_r = \sqrt{225 + 25}$

$V_r = \sqrt{250}$

$V_r = 15.81 \, \text{km/h}$

Key Concept: Basic Concept

d) 9.8 m/s²

[Solution Description] The acceleration due to gravity near the Earth’s surface in free fall is a constant value and it is denoted by $g$. According to the given syllabus, $g$ is $9.8 \, \text{m/s}^2$.

Key Concept: Real-World Application

b) 4 m/s$^2$

[Solution Description] Use the formula $a = \frac{v – v_0}{t}$. Here, $v = 30$ m/s, $v_0 = 10$ m/s, and $t = 5$ s. Thus:

$a = \frac{30 – 10}{5} = \frac{20}{5} = 4 \, \text{m/s}^2.$

Key Concept: Graph Interpretation

b) Constant velocity

[Solution Description] A velocity-time graph that is a straight horizontal line indicates constant velocity. There is no change in velocity over time, meaning acceleration is zero.

Key Concept: Definition Check

c) The velocity at a specific instant

[Solution Description] Instantaneous velocity is defined as the limit of average velocity as the time interval approaches zero. It specifically refers to the velocity at a particular moment in time, and it can be represented by the slope of the tangent line on a position-time graph.

Key Concept: Complex Calculus Problem, Multi-Step Problem

c) 213 m/s

[Solution Description] To find the instantaneous velocity, we need to differentiate the position function with respect to time $t$. The derivative of $x(t)$ gives us the velocity function $v(t)$.

First, differentiate each term:

$\frac{dx}{dt} = \frac{d}{dt}(8t^4) – \frac{d}{dt}(5t^3) + \frac{d}{dt}(6t^2) – \frac{d}{dt}(7t) + \frac{d}{dt}(9).$

Calculating each derivative separately:

$\frac{d}{dt}(8t^4) = 32t^3,$

$\frac{d}{dt}(5t^3) = 15t^2,$

$\frac{d}{dt}(6t^2) = 12t$,

$\frac{d}{dt}(7t) = 7$,

$\frac{d}{dt}(9) = 0$.

Thus, the velocity function becomes:

$v(t) = 32t^3 – 15t^2 + 12t – 7.$

Substitute $t = 2$ into the derived velocity function:

$v(2) = 32(2)^3 – 15(2)^2 + 12(2) – 7$

$= 32 \times 8 – 15 \times 4 + 24 – 7$

$= 256 – 60 + 24 – 7$

$= 213 \, \text{m/s}$.

Therefore, the instantaneous velocity at $t = 2$ s is 213 m/s.

Key Concept: Calculating Instantaneous Velocity

c) 26 m/s

[Solution Description] To find the instantaneous velocity, we need to differentiate the position function $x(t)$ with respect to time $t$. The derivative $v(t) = \frac{dx}{dt}$ represents the velocity function.

$\frac{dx}{dt} = \frac{d}{dt}(4t^2 + 2t) = 8t + 2.$

Now, substitute$t = 3$ into this velocity function:

$v(3) = 8(3) + 2 = 24 + 2 = 26 \, \text{m/s}.$

Therefore, the instantaneous velocity at $t = 3$ seconds is $26 \, \text{m/s}$.

Key Concept: Position-Time Graph Analysis

c) Increasing velocity

[Solution Description] The slope of a tangent on a position-time graph represents the instantaneous velocity. An increasing slope implies an increase in the steepness of the graph over time, indicating that the object’s velocity is increasing.

Key Concept: Simple Calculation

b) 72 km/h

[Solution Description] To calculate the average speed, we use the formula:

$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}$

Let’s assume the distance between city A and city B is d. The total distance for the round trip is $2d$. The time taken to travel from A to B is $\frac{d}{90}$ hours, and the time taken to return is $\frac{d}{60}$ hours.

$\text{Total time} = \frac{d}{90} + \frac{d}{60} = \frac{2d}{180} + \frac{3d}{180} = \frac{5d}{180} = \frac{d}{36}$

$\text{Average speed} = \frac{2d}{\frac{d}{36}} = \frac{2d \times 36}{d} = 72 \, \text{km/h}$

So, the average speed for the entire journey is 72 km/h.

Key Concept: Average Acceleration Formula

d) $$a = \frac{\Delta v}{\Delta t}$$

[Solution Description]

The formula for average acceleration is given by $$a = \frac{\Delta v}{\Delta t}$$ where $$\Delta v$$ is the change in velocity and $$\Delta t$$ is the change in time.

Key Concept: Free Fall Concept

c) $v = v_0 – gt$

[Solution Description]

The equation for the velocity of an object under free fall is given by:

$v = v_0 – gt$

where $v$ is the final velocity, $v_0$ is the initial velocity, $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$), and $t$ is the time.

Hence, the correct expression is $v = v_0 – gt$.

Key Concept: Intermediate Calculation

b) 8 m/s

[Solution Description] To find the velocity of the boat with respect to the riverbank, we need to add the velocities of the boat and the river since they are moving in the same direction.

The velocity of the boat relative to the riverbank is given by:

$$v_{\text{boat/ground}} = v_{\text{boat/water}} + v_{\text{water/ground}}$$

Substituting the values, we have:

$$v_{\text{boat/ground}} = 5 \, \text{m/s} + 3 \, \text{m/s} = 8 \, \text{m/s}$$

Hence, the velocity of the boat with respect to the riverbank is 8 m/s.

Key Concept: Basic Calculation

d) -29.4 m/s

[Solution Description]

Using the kinematic equation $v = v_0 – gt$, where the initial velocity $v_0 = 0$, and time $t = 3$ seconds, we calculate:

$v = 0 – (9.8 \, \text{m/s}^2) \times 3 = -29.4 \, \text{m/s}.$

Therefore, the final velocity is $-29.4 \, \text{m/s}$.

Key Concept: Complex Calculus, Advanced Reasoning

c) 237 m/s

[Solution Description] To find the instantaneous velocity, we first differentiate the position function with respect to time $t$:

$v(t) = \frac{dx}{dt} = \frac{d}{dt}(5t^3 – 3t + 7) = 15t^2 – 3$

Now calculate $v(4)$:

$v(4) = 15(4)^2 – 3 = 240 – 3 = 237\, \text{m/s}$

Key Concept: Velocity-Time Graph Slope

c) Acceleration

[Solution Description]

The slope of a velocity-time graph is given by $\frac{\Delta v}{\Delta t}$, where $\Delta v$ is the change in velocity and $\Delta t$ is the change in time. By definition, this represents acceleration.

Key Concept: Definition Recall

c) Change in position with time

[Solution Description] Motion is described in the syllabus as a change in position with time. Therefore, the correct definition aligns with this description.

Key Concept: Relative Velocity Application

a) 20 km/h

[Solution Description] The formula for relative velocity $$v_{\text{relative}} = v_1 – v_2$$ is used here, where $$v_1 = 60$$ km/h (velocity of the car) and $$v_2 = 80$$ km/h (velocity of the train). Therefore,

$$v_{\text{relative}} = 60\, \text{km/h} – 80\, \text{km/h} = -20\, \text{km/h}.$$

The negative sign indicates that the car is moving slower than the train. Thus, the velocity of the car relative to the train is 20 km/h.

Key Concept: Graph Interpretation

a) Straight line parallel to time axis

[Solution Description] In uniform motion, there is no change in velocity; hence, the acceleration is zero. This means that on a velocity-time graph, the line representing the motion will be a straight line parallel to the time axis.