A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t seconds, the total distance traveled is :
A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t seconds, the total distance traveled is :
Option 1 -
(4αβ / (α + β)) * t²
Option 2 -
(αβ / 4(α + β)) * t²
Option 3 -
(2αβ / (α + β)) * t²
Option 4 -
(αβ / 2(α + β)) * t²
Asked by Shiksha User
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1 Answer
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Correct Option - 4
Detailed Solution:t? + t? = t
v? /t? = α and v? /t? = β
t? = v? /α and t? = v? /β
v? /α + v? /β = t
v? (1/α + 1/β) = t
v? = (αβt) / (α + β)
Distance traveled = Area under speed-time graph
Distance = ½ × base × height = ½ × t × v?
Distance = ½ × t × (αβt) / (α + β) = (αβt²) / (2 (α + β)
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