Chapter 2: Motion Along a Straight Line

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Motion along a straight line, or kinematics in one dimension, is a foundational topic in physics, describing how objects move without considering the forces causing that motion. This chapter explores displacement and distance, velocity, acceleration, kinematic equations for constant acceleration, and free fall, essential for JEE Main, JEE Advanced, and NEET. These concepts appear in problems on motion of vehicles, falling objects, and relative motion. With detailed derivations, exam-focused examples, and robust problem-solving strategies, this chapter equips students to master kinematics for competitive exams, ensuring precision and conceptual clarity.

2.1 Displacement, Distance, and Position

Displacement is a vector quantity, defined as the change in position, $\mathrm{\Delta }x=x_f-x_i$, where $x_f$ is the final position and $x_i$ is the initial position, measured in meters ($m$). Distance is the scalar total path length traveled, always positive. For example, a car moving 5 m east then 3 m west has a distance of 8 m but a displacement of 2 m east. Position ($x$) is the location relative to a reference point (origin). JEE/NEET problems often involve distinguishing displacement from distance and calculating position. Fundamentals of Physics emphasizes the vector nature of displacement.

Derivation: Displacement in Terms of Velocity
For constant velocity $v$, displacement is $\mathrm{\Delta }x=vt$, where $t$ is time. If velocity varies, use the average velocity:

$$v_{\text{avg}}=\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}\text{so}\mathrm{\Delta }x=v_{\text{avg}}\mathrm{\Delta }t$$

For non-constant velocity, instantaneous velocity $v=\frac{dx}{dt}$ requires integration:

$$\mathrm{\Delta }x={\int }_{t_i}^{t_f}v(t)dt$$

Solved Example: A JEE Main problem involves a car moving 10 m east, then 6 m west. Calculate the displacement and distance traveled.

• Solution:
  Distance: Total path length = $10+6=16m$.
  Displacement: $\mathrm{\Delta }x=10\text{(east)}-6\text{(west)}=4m$ east (2 significant figures, April 14, 2025).
  • JEE Tip: Displacement considers direction; distance does not. Common error: Ignoring direction in displacement.

Solved Example: A NEET problem asks for the displacement of a particle with position $x=5t^2-2t+3$ from $t=1s$ to $t=2s$.

• Solution:
  Position at $t=1$: $x(1)=5(1{)}^2-2(1)+3=6m$.
  Position at $t=2$: $x(2)=5(2{)}^2-2(2)+3=19m$.
  Displacement:$$\mathrm{\Delta }x=x(2)-x(1)=19-6=13m$$
  • NEET Tip: Use the position function directly for displacement. Common error: Confusing displacement with distance.

Solved Example: A JEE Advanced problem involves a particle moving with velocity $v=3t^2-6t$. Find the displacement from $t=0s$ to $t=2s$.

• Solution:
  Displacement is the integral of velocity:$$\mathrm{\Delta }x={\int }_0^{2}v(t)dt={\int }_0^2(3t^2-6t)dt$$Integrate: $3t^2-6t$ becomes $t^3-3t^2$. Evaluate:$$\mathrm{\Delta }x=[t^3-3t^2{]}_0^2=(2^3-3\cdot 2^2)-(0-0)=8-12=-4m$$
  • JEE Tip: Negative displacement indicates direction opposite to the reference. Common error: Forgetting to evaluate at limits.

Application: Displacement is used in navigation (e.g., GPS tracking a car’s net movement), while distance informs fuel consumption calculations.

2.2 Average and Instantaneous Velocity

Average velocity is a vector, defined as $v_{\text{avg}}=\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}$, with units $m/s$. Instantaneous velocity is the velocity at a specific moment, $v=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}=\frac{dx}{dt}$. For constant velocity, $v_{\text{avg}}=v$. JEE/NEET problems involve calculating average velocity from displacement and instantaneous velocity from position functions. University Physics highlights the distinction between average and instantaneous quantities.

Derivation: Instantaneous Velocity from Position
If position is $x(t)$, instantaneous velocity is the derivative:

$$v=\frac{dx}{dt}$$

For example, if $x=4t^2-3t+2$, then:

$$v=\frac{d}{dt}(4t^2-3t+2)=8t-3$$

Average velocity over an interval requires displacement: $v_{\text{avg}}=\frac{x(t_2)-x(t_1)}{t_2-t_1}$.

Solved Example: A JEE Main problem involves a car traveling 100 m in 5 s, then 50 m in 5 s in the same direction. Calculate the average velocity.

• Solution:
  Total displacement: $\mathrm{\Delta }x=100+50=150m$.
  Total time: $\mathrm{\Delta }t=5+5=10s$.
  Average velocity:$$v_{\text{avg}}=\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}=\frac{150}{10}=15m/s$$
  • JEE Tip: Use total displacement for average velocity. Common error: Averaging velocities directly.

Solved Example: A NEET problem gives a particle’s position as $x=2t^3-5t^2+3$. Find the instantaneous velocity at $t=3s$.

• Solution:
  Velocity: $v=\frac{dx}{dt}$. Given $x=2t^3-5t^2+3$:$$v=\frac{d}{dt}(2t^3-5t^2+3)=6t^2-10t$$At $t=3$: $v=6(3{)}^2-10(3)=54-30=24m/s$.
  • NEET Tip: Differentiate position to find velocity. Common error: Using average velocity formula.

Solved Example: A JEE Advanced problem involves a particle with velocity $v=4t-2$. Find the average velocity from $t=1s$ to $t=3s$.

• Solution:
  Average velocity: $v_{\text{avg}}=\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}$, where $\mathrm{\Delta }x={\int }_1^{3}v(t)dt$.$$\mathrm{\Delta }x={\int }_1^3(4t-2)dt=[2t^2-2t{]}_1^3=(2(3{)}^2-2(3))-(2(1{)}^2-2(1))=(18-6)-(2-2)=12m$$$\mathrm{\Delta }t=3-1=2s$. Thus:$$v_{\text{avg}}=\frac{12}{2}=6m/s$$
  • JEE Tip: Integrate velocity to find displacement for average velocity. Common error: Averaging velocities at endpoints.

Application: Average velocity helps calculate travel times (e.g., a train’s journey), while instantaneous velocity informs speedometer readings.

2.3 Average and Instantaneous Acceleration

Average acceleration is a vector, defined as $a_{\text{avg}}=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$, with units $m/s^2$. Instantaneous acceleration is $a=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}=\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}$, the second derivative of position. For constant acceleration, $a_{\text{avg}}=a$. JEE/NEET problems involve calculating acceleration from velocity or position functions. Fundamentals of Physics emphasizes acceleration’s role in motion dynamics.

Derivation: Instantaneous Acceleration from Velocity
If velocity is $v(t)$, acceleration is:

$$a=\frac{dv}{dt}$$

If $v=3t^2-4t+1$, then:

$$a=\frac{d}{dt}(3t^2-4t+1)=6t-4$$

If position is $x(t)$, acceleration is: $a=\frac{d^{2}x}{d{t}^2}$. For $x=t^3-2t^2+t$:
$v=\frac{dx}{dt}=3t^2-4t+1$, then:

$$a=\frac{dv}{dt}=6t-4$$

Solved Example: A JEE Main problem involves a car with velocity changing from $10m/s$ to $20m/s$ in $4s$. Calculate the average acceleration.

• Solution:
  $\mathrm{\Delta }v=20-10=10m/s$, $\mathrm{\Delta }t=4s$. Average acceleration:$$a_{\text{avg}}=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}=\frac{10}{4}=2.5m/s^2$$(2 significant figures).
  • JEE Tip: Use change in velocity for average acceleration. Common error: Ignoring direction.

Solved Example: A NEET problem gives a particle’s velocity as $v=5t^2-3t+2$. Find the instantaneous acceleration at $t=2s$.

• Solution:
  Acceleration: $a=\frac{dv}{dt}$. Given $v=5t^2-3t+2$:$$a=\frac{d}{dt}(5t^2-3t+2)=10t-3$$At $t=2$: $a=10(2)-3=17m/s^2$.
  • NEET Tip: Differentiate velocity to find acceleration. Common error: Using average acceleration formula.

Solved Example: A JEE Advanced problem involves a particle with position $x=2t^3-6t^2+4t$. Find the instantaneous acceleration at $t=1s$.

• Solution:
  Velocity: $v=\frac{dx}{dt}=\frac{d}{dt}(2t^3-6t^2+4t)=6t^2-12t+4$.
  Acceleration:$$a=\frac{dv}{dt}=\frac{d}{dt}(6t^2-12t+4)=12t-12$$At $t=1$: $a=12(1)-12=0m/s^2$.
  • JEE Tip: Acceleration can be zero even if velocity is not. Common error: Forgetting the second derivative.

Application: Acceleration is key in vehicle design (e.g., car braking systems) and sports (e.g., a sprinter’s start).

2.4 Kinematic Equations and Free Fall

Kinematic equations describe motion with constant acceleration:

1. $v=u+at$
2. $x=ut+\frac{1}{2}a{t}^2$ (displacement from initial position)
3. $v^2=u^2+2ax$
4. $x=\frac{1}{2}(u+v)t$
   where $u$ is initial velocity, $v$ is final velocity, $a$ is acceleration, $t$ is time, and $x$ is displacement. Free fall is motion under gravity ($g\approx 9.8m/s^2$), with $a=-g$ (upward positive). JEE/NEET problems involve solving for unknowns using these equations. HC Verma emphasizes their application in free fall and projectile motion.

Derivation: Kinematic Equation $v=u+at$
For constant acceleration $a$, acceleration is: $a=\frac{dv}{dt}$. Integrate:

$${\int }_u^{v}dv={\int }_0^{t}adt$$

Since $a$ is constant:

$$v-u=at\text{so}v=u+at$$

Derivation: Kinematic Equation $x=ut+\frac{1}{2}a{t}^2$
Velocity is $v=u+at$. Displacement: $x={\int }_0^{t}vdt$. Substitute $v$:

$$x={\int }_0^t(u+at)dt=ut+\frac{1}{2}a{t}^2$$

Solved Example: A JEE Main problem involves a car starting from rest with $a=2m/s^2$. Find the velocity after $5s$.

• Solution:
  Use $v=u+at$. Given $u=0$, $a=2m/s^2$, $t=5s$:$$v=0+2\times 5=10m/s$$
  • JEE Tip: Identify knowns and choose the right equation. Common error: Forgetting initial velocity.

Solved Example: A NEET problem involves a ball dropped from a height of $20m$ ($g=9.8m/s^2$). Find the time to reach the ground.

• Solution:
  Use $x=ut+\frac{1}{2}a{t}^2$. Given $x=-20m$ (downward), $u=0$, $a=-g=-9.8m/s^2$:$$-20=0+\frac{1}{2}(-9.8)t^2\Rightarrow -20=-4.9t^2\Rightarrow t^2=\frac{20}{4.9}\approx 4.08\Rightarrow t\approx 2.02s$$Round to 2 significant figures: $2.0s$.
  • NEET Tip: Use $g$ downward as negative. Common error: Incorrect sign for displacement.

Solved Example: A JEE Advanced problem involves a car with $u=10m/s$, $a=-2m/s^2$. Find the displacement after $4s$.

• Solution:
  Use $x=ut+\frac{1}{2}a{t}^2$. Given $u=10m/s$, $a=-2m/s^2$, $t=4s$:$$x=(10)(4)+\frac{1}{2}(-2)(4{)}^2=40-\frac{1}{2}\cdot 2\cdot 16=40-16=24m$$
  • JEE Tip: Account for deceleration with negative $a$. Common error: Forgetting the $\frac{1}{2}$ factor.

Solved Example: A JEE Main problem involves a ball thrown upward with $u=15m/s$ ($g=9.8m/s^2$). Find the time to reach the maximum height.

• Solution:
  At maximum height, $v=0$. Use $v=u+at$, with $u=15m/s$, $a=-g=-9.8m/s^2$:$$0=15+(-9.8)t\Rightarrow 9.8t=15\Rightarrow t=\frac{15}{9.8}\approx 1.53s$$Round to 2 significant figures: $1.5s$.
  • JEE Tip: Set final velocity to zero at peak. Common error: Using positive $g$.

Application: Kinematic equations model car braking distances, while free fall applies to skydiving safety calculations.

Summary and Quick Revision

• Displacement and Distance: Displacement: $\mathrm{\Delta }x=x_f-x_i$ (vector, $m$). Distance: Total path length (scalar, $m$).
• Velocity: Average: $v_{\text{avg}}=\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}$. Instantaneous: $v=\frac{dx}{dt}$. Units: $m/s$.
• Acceleration: Average: $a_{\text{avg}}=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$. Instantaneous: $a=\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}$. Units: $m/s^2$.
• Kinematic Equations: For constant $a$:
  • $v=u+at$
  • $x=ut+\frac{1}{2}a{t}^2$
  • $v^2=u^2+2ax$
  • $x=\frac{1}{2}(u+v)t$
• Free Fall: Motion under gravity, $a=-g\approx -9.8m/s^2$ (upward positive).
• SI Units: Displacement ($m$), velocity ($m/s$), acceleration ($m/s^2$), time ($s$).
• JEE/NEET Tips: Use correct signs for direction, verify significant figures (April 14, 2025), choose the right kinematic equation, integrate for non-constant motion.
• Applications: Car braking, skydiving, sports, navigation.

Practice Problems

Explore our extensive problem set with 100 problems inspired by JEE Main, JEE Advanced, and NEET patterns to test your understanding of motion along a straight line.

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Note: Content regularly updated to align with current JEE/NEET syllabi.