A car braking uniformly decelerates from 20 m/s to rest with deceleration 5 m/s². The braking distance is:
A20 m, equal to the initial velocity numerically
B40 m, equal to the velocity divided by deceleration
C40 m, computed from v² = u² + 2as as u²/(2a)
D80 m, equal to the velocity squared divided by 2
Answer & Solution
Correct answer: C. 40 m, computed from v² = u² + 2as as u²/(2a)
v² = u² + 2as → 0 = 20² + 2(-5)s → 0 = 400 - 10s → s = 40 m.
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