JEE PYQ: Motion in a Straight Line - Question ID 199346dbcaa3 (JEE Main 2024)

Core Formula & Concept:

When a body moves with uniform (constant) acceleration or retardation, its motion can be described using the three fundamental kinematic equations for straight-line motion. In this problem, the bus is decelerating uniformly (retardation is constant) until it comes to rest. The key formulas we use are:

• $v = u + at$
  where:
  • $v$ = final velocity (0 m/s, since the bus halts),
  • $u$ = initial velocity (given in km/h, must be converted to m/s),
  • $a$ = acceleration (negative for retardation),
  • $t$ = time taken (4 s).
• $s = ut + \frac{1}{2}at^2$
  where $s$ is the distance travelled during the braking period.

The concept of uniform retardation implies that the acceleration $a$ is constant and negative (opposite to the direction of motion). We first determine $a$ using the first equation, then substitute it into the second to find $s$.

Step-by-Step Derivation:

Step 1: Convert the initial speed to SI units

The bus speed is given as $72 \mathrm{~km/h}$. To convert to $\mathrm{m/s}$: $72 \frac{\mathrm{km}}{\mathrm{h}} = 72 \times \frac{1000 \mathrm{~m}}{3600 \mathrm{~s}} = 20 \mathrm{~m/s}.$ Thus, $u = 20 \mathrm{~m/s}$.

Step 2: Determine the acceleration (retardation)

The bus comes to rest, so $v = 0 \mathrm{~m/s}$. Using $v = u + at$: $0 = 20 + a \times 4 \implies a = \frac{-20}{4} = -5 \mathrm{~m/s^2}.$ The negative sign indicates retardation (deceleration).

Step 3: Calculate the distance travelled during braking

Using $s = ut + \frac{1}{2}at^2$: $s = 20 \times 4 + \frac{1}{2} \times (-5) \times (4)^2 = 80 + \frac{1}{2} \times (-5) \times 16 = 80 - 40 = 40 \mathrm{~m}.$

Step 4: Verify using an alternative formula

We can also use $v^2 = u^2 + 2as$: $0 = (20)^2 + 2 \times (-5) \times s \implies 0 = 400 - 10s \implies s = \frac{400}{10} = 40 \mathrm{~m}.$ This confirms our previous result.

Common Traps & Exam Tip:

Unit Conversion Error: Many students forget to convert km/h to m/s, leading to incorrect values for $u$ and consequently wrong $a$ and $s$. Always ensure units are consistent (SI units are preferred in kinematics).

Sign of Acceleration: Students often confuse the sign of $a$. Since the bus is slowing down, $a$ must be negative. Using $a = +5 \mathrm{~m/s^2}$ would yield $s = 120 \mathrm{~m}$, which is incorrect.

Direct Use of Average Velocity: Another valid approach is using average velocity: $s = \frac{u + v}{2} \times t = \frac{20 + 0}{2} \times 4 = 40 \mathrm{~m}$. This is a quick sanity check.

Exam Tip: Always cross-verify using at least two different kinematic equations to ensure consistency and catch calculation errors.