(A) Let the uniform acceleration be $a$. Using the third equation of motion $v^2 = u^2 + 2as$: For the interval $AB$: $v_B^2 = v_A^2 + 2a(AB)$ $6^2 =

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(A) Let the uniform acceleration be $a$. Using the third equation of motion $v^2 = u^2 + 2as$: For the interval $AB$: $v_B^2 = v_A^2 + 2a(AB)$ $6^2 = 3^2 + 2a(60)$ $36 = 9 + 120a$ $27 = 120a$ $a = 27/120 = 9/40 \text{ m/s}^2$. For the interval $BC$: $v_C^2 = v_B^2 + 2a(BC)$ $9^2 = 6^2 + 2(9/40)(BC)$ $81 = 36 + (9/20)(BC)$ $45 = (9/20)(BC)$ $BC = 45 \times 20 / 9 = 5 \times 20 = 100 \text{ m}$.

Class 11 Physics Motion in a Straight Line Uniformly Accelerated Motion

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$A$ car passes three points $A$, $B$ and $C$ at $3 \text{ m/s}$, $6 \text{ m/s}$ and $9 \text{ m/s}$ respectively in a straight line with uniform acceleration. If distance $AB = 60 \text{ m}$, then find the distance $BC$. (in $m$)

MHT CET 2026 All MHT CET

A
$100$
B
$75$
C
$200$
D
$125$

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Motion in a Straight Line

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Uniformly Accelerated Motion

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$A$ car passes three points $A$, $B$ and $C$ at $3 \text{ m/s}$, $6 \text{ m/s}$ and $9 \text{ m/s}$ respectively in a straight line with uniform acceleration. If distance $AB = 60 \text{ m}$, then find the distance $BC$. (in $m$)

Similar Questions

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