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A car is moving with uniform acceleration. It covers 200 m in 2 sec and 220 m in next 4 sec. The velocity of car at the end of 7th second from start is
Options
(a) 1 m/s
(b) 0.1 m/s
(c) 10 m/s
(d) 20 m/s
Correct Answer:
10 m/s
Explanation:
For first 2 second
S = 2u +1/2α (2)²
200 = 2u + 2α
or u + α = 100 …(i)
For first 2 + 4 = 6 second
S = 200 + 220 = 420 m
420 = 6u + 36/2 α
420 = 6u + 18α ⇒ 420 = 6(u + 3α)
u + 3α = 420/6 = 70 …(ii)
Solving equation (i) and (ii)
α = -15 m/sec²
This is the retardation.
Using these formula V=ut+1/2 αt²
200 = u+1/2 (-15)(2)²
u = 115
Hence velocity after 7 second is
V = u+αt
= 115 + (–15) x 7
= 10 m/sec.
Related Questions: - The resistance of a bulb filament is 100 Ω at a temperature of 100⁰C.
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Topics: Motion in Straight Line
(93)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The resistance of a bulb filament is 100 Ω at a temperature of 100⁰C.
- The coherence of two light sources means that the light waves emitted have
- The Davisson- Germer experiment is the direct evidence of
- A carnot’s engine operates with source at 127° C and sink at 27°C
- The equation of state for 5 g of oxygen at a pressure P and temperature T,
Topics: Motion in Straight Line (93)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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