Q103 / 150 · MHT CET 2026 (11 April Shift 1) · Physics
A car passes three points A, B and C at $3$ m/s, $6$ m/s and $9$ m/s respectively in a straight line with uniform acceleration. If distance $AB = 60$ m then find the distance BC.
- Correct answerA$100$ m
- B$75$ m
- C$200$ m
- D$125$ m
Solution
Using the equation of motion $v^2 - u^2 = 2as$ for the interval AB:
$v_B^2 - v_A^2 = 2a(AB)$
$6^2 - 3^2 = 2a(60)$
$36 - 9 = 120a$
$27 = 120a \Rightarrow a = \dfrac{27}{120} = \dfrac{9}{40}$ m/s$^2$
Applying the same equation for the interval BC:
$v_C^2 - v_B^2 = 2a(BC)$
$9^2 - 6^2 = 2 \times \dfrac{9}{40} \times BC$
$81 - 36 = \dfrac{9}{20} \times BC$
$45 = \dfrac{9}{20} \times BC$
$BC = \dfrac{45 \times 20}{9} = 100$ m
Answer: $100$ m
$v_B^2 - v_A^2 = 2a(AB)$
$6^2 - 3^2 = 2a(60)$
$36 - 9 = 120a$
$27 = 120a \Rightarrow a = \dfrac{27}{120} = \dfrac{9}{40}$ m/s$^2$
Applying the same equation for the interval BC:
$v_C^2 - v_B^2 = 2a(BC)$
$9^2 - 6^2 = 2 \times \dfrac{9}{40} \times BC$
$81 - 36 = \dfrac{9}{20} \times BC$
$45 = \dfrac{9}{20} \times BC$
$BC = \dfrac{45 \times 20}{9} = 100$ m
Answer: $100$ m
Hence, the correct answer is option: 'A'.