Question

in a car race,car A takes a time t less than car B at the finish point and passes the finishing point with speed v more than that of car B. Assuming that both cars starts from rest and travel with constant acceleration a1 and a2 respectively. show that v=
$\sqrt{a1a2t}$
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Answer 1

Answer:

Explanation:

Time taken by a = t1

Time taken by b = t2

t = t2 - t1

v = v1 - v2

$v^{2} - u^{2} = 2as$

$v_{1} ^{2} - 0 = 2a_{1} ^{2} s$

$v_{2} ^{2} - 0 = 2a_{2} ^{2} s$

$v_{1} = \sqrt{2a_{1}s }$

$v_{2} = \sqrt{2a_{2}s }$

t1 = $\sqrt{\frac{2s}{a1} }$

t2 = $\sqrt{\frac{2s}{a2} }$

v/t = v1 - v2/t2 - t1

$\frac{v}{t} = \frac{v_{1} - v_{2}}{t_{2} - t_{1} }$

    = $\frac{{\sqrt{2a1s} - \sqrt{2a2s} } }{\sqrt{\frac{2s}{a2}} - \sqrt{\frac{2s}{a1} } }$

under root 2s gets eliminated

=$\frac{\sqrt{a1} - \sqrt{a2} }{\frac{\sqrt{a1} - \sqrt{a2} }{\sqrt{a1a2} } }$

under root a1 - under root a2 gets eliminated

v = t $\sqrt{a1a2}$