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 - <p>(4αβ / (α + β)) * t²<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>(αβ / 4(α + β)) * t²<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>(2αβ / (α + β)) * t²<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>(αβ / 2(α + β)) * t²</p>
4 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
V
5 months ago
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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Physics Motion in Straight Line 2025

Physics Motion in Straight Line 2025

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