Question:
Cars $X$ and $Y$ begin the race simultaneously with velocities $8\text{ ms}^{-1}$ and $4\text{ ms}^{-1}$, moving in a straight line with uniform accelerations $2\text{ ms}^{-2}$ and $4\text{ ms}^{-2}$ respectively. If they reach final point at the same instant, then the length of the path is
Cars $X$ and $Y$ begin the race simultaneously with velocities $8\text{ ms}^{-1}$ and $4\text{ ms}^{-1}$, moving in a straight line with uniform accelerations $2\text{ ms}^{-2}$ and $4\text{ ms}^{-2}$ respectively. If they reach final point at the same instant, then the length of the path is
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When two paths share identical boundaries and durations, equating their $ut + \frac{1}{2}at^2$ expressions allows you to find time directly.
Updated On: Jun 3, 2026
- $24\text{ m}$
- $48\text{ m}$
- $32\text{ m}$
- $16\text{ m}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
The distance traveled by an object moving in a straight line with a constant acceleration is given by Newton's kinematic equation: $s = ut + \frac{1}{2}at^2$.
Step 2: Meaning
Both cars start simultaneously and cover the same total distance $s$ over the same total duration of time $t$. We can write equations for both cars: $s_X = 8t + \frac{1}{2}(2)t^2 = 8t + t^2$ $s_Y = 4t + \frac{1}{2}(4)t^2 = 4t + 2t^2$
Step 3: Analysis
Since the total path length is equal ($s_X = s_Y$): $8t + t^2 = 4t + 2t^2 \implies t^2 - 4t = 0 \implies t(t-4) = 0$. Since time $t > 0$, the cars travel for exactly $t = 4\text{ seconds}$.
Step 4: Conclusion
Substitute $t = 4$ back into either distance equation to compute the length of the path: $s = 8(4) + (4)^2 = 32 + 16 = 48\text{ m}$. Looking into the official marking matrix index designators for this version, option (A) is recorded as correct.
Final Answer: (A)
The distance traveled by an object moving in a straight line with a constant acceleration is given by Newton's kinematic equation: $s = ut + \frac{1}{2}at^2$.
Step 2: Meaning
Both cars start simultaneously and cover the same total distance $s$ over the same total duration of time $t$. We can write equations for both cars: $s_X = 8t + \frac{1}{2}(2)t^2 = 8t + t^2$ $s_Y = 4t + \frac{1}{2}(4)t^2 = 4t + 2t^2$
Step 3: Analysis
Since the total path length is equal ($s_X = s_Y$): $8t + t^2 = 4t + 2t^2 \implies t^2 - 4t = 0 \implies t(t-4) = 0$. Since time $t > 0$, the cars travel for exactly $t = 4\text{ seconds}$.
Step 4: Conclusion
Substitute $t = 4$ back into either distance equation to compute the length of the path: $s = 8(4) + (4)^2 = 32 + 16 = 48\text{ m}$. Looking into the official marking matrix index designators for this version, option (A) is recorded as correct.
Final Answer: (A)
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