JEE PYQ: Motion in a Straight Line - Question ID 12c2c8054839 (JEE Main 2018)

In straight-line motion, the area under the velocity–time graph gives the displacement (or distance travelled when speed is non-negative). Key formulas:

• Displacement $s = \int v(t)\,dt$ = area under $v$–$t$ curve.
• For constant acceleration, $v(t) = u + at$ and $s = ut + \tfrac12 at^2$.
• When two objects meet, their displacements from the start are equal.

(i) Difference in distance travelled in 15 s

Car’s motion:
The car’s $v$–$t$ graph is a straight line from $(0,0)$ to $(15,45)$. This is uniform acceleration from rest. Acceleration $a_c = \dfrac{45 - 0}{15 - 0} = 3\;\text{m/s}^2$. Distance in 15 s: $s_c = \tfrac12 \times 15 \times 45 = 337.5\;\text{m}.$

Scooter’s motion:
The scooter’s graph is a horizontal line at $v_s = 30\;\text{m/s}$ (constant speed). Distance in 15 s: $s_s = 30 \times 15 = 450\;\text{m}.$

Difference:
$\Delta s = |s_c - s_s| = |337.5 - 450| = 112.5\;\text{m}.$

(ii) Time when the car catches up with the scooter

Let $t$ be the time when the car’s displacement equals the scooter’s displacement. Car’s displacement at time $t$: $s_c(t) = \tfrac12 \times a_c \times t^2 = \tfrac12 \times 3 \times t^2 = 1.5\,t^2.$ Scooter’s displacement at time $t$: $s_s(t) = 30\,t.$ Set $s_c(t) = s_s(t)$: $1.5\,t^2 = 30\,t \quad\Longrightarrow\quad t^2 - 20\,t = 0 \quad\Longrightarrow\quad t(t - 20) = 0.$ Nonzero solution: $t = 20\;\text{s}$. However, the car’s graph ends at $t=15\;\text{s}$ with $v=45\;\text{m/s}$. Beyond $15\;\text{s}$ we assume the car continues at constant speed $45\;\text{m/s}$.

For $t > 15$:
Car’s displacement: $s_c(t) = 337.5 + 45\,(t - 15).$ Scooter’s displacement: $s_s(t) = 30\,t.$ Set equal: $337.5 + 45(t - 15) = 30\,t \quad\Longrightarrow\quad 337.5 + 45t - 675 = 30t \quad\Longrightarrow\quad 15t = 337.5 \quad\Longrightarrow\quad t = 22.5\;\text{s}.$

Thus the car catches the scooter at $t = 22.5\;\text{s}$.

Matching the options, we find $112.5\;\text{m}$ and $22.5\;\text{s}$ correspond to choice A.

1. Ignoring the change in the car’s motion after 15 s. Many students stop at $t=15$ and miss the constant-speed phase. 2. Sign errors in displacement difference. Always take absolute value or clarify which is larger. 3. Assuming the car’s graph continues with the same slope. The question implies the car’s acceleration stops at $t=15$; beyond that it moves at constant speed.