Home › CBSE Class 11 › Physics › Motion in a Straight Line › Per NCERT Example 2.6, the STOPPING DISTANCE of …
Per NCERT Example 2.6, the STOPPING DISTANCE of a vehicle (with initial velocity v₀ and deceleration −a) is given by which expression?
Ad_s = v₀² × a
Bd_s = 2v₀/a
Cd_s = v₀² / (2a) — proportional to the square of the initial velocity
Dd_s = v₀/a
Answer & Solution
Correct answer: C. d_s = v₀² / (2a) — proportional to the square of the initial velocity
Per NCERT Example 2.6, using v² = v₀² + 2ax with v = 0: x = v₀²/(2a) (with sign convention).
Related questions
A ball is thrown up with initial velocity $u$. It reaches its highest point at time:For a ball thrown vertically upwards and returning to the ground, the velocity-time graph:Displacement differs from distance in that displacement:For uniform motion, velocity is represented by the gradient of a:Which quantities need a sign convention fixed before they are assigned values?In uniform motion, how does instantaneous velocity vary with time?Why can average speed exceed the magnitude of average velocity?Speed differs from velocity in that speed does not carry: