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Motion in a Straight Line — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Straight Line3 Marks
Question
A car starting from rest, accelerates at the rate f through a distances, then continues at constant speed for sometimet and then decelerate at the rate $\frac{\text{f}}{2}$ to come to rest. If the total distance is $5s$, then prove that $\text{s}=\frac{1}{2}\text{ft}^2.$
✓

Answer

For accelerated motion, $\text{u}=0, \text{a}=\text{f}, \text{s}=\text{s}$ As $\text{v}^2-\text{u}^2=2\text{as},$
$\therefore \text{v}^2_1-0^2=2\text{fs}$
$\Rightarrow\text{v}_1=\sqrt{2\text{fs}},$ Distance travelled, $\text{s}_2=\text{v}_1\text{t}=\text{t}\sqrt{2\text{fs}}$ For decelerated motion, $\text{u}=\sqrt{2\text{fs}},\text{a}=-\frac{\text{f}}{2},\text{v}=0$ As $\text{v}^2-\text{u}^2=2\text{as}$
$\therefore 0^2-(\sqrt{2\text{fs}})^2=2\times\Big(-\frac{\text{f}}{2}\Big)\text{s}_3$ Distance travelled $s_3​​​​​​​ = 2s$ Given, $\text{s}+\text{s}_2+\text{s}_3=5\text{s}$
$\Rightarrow \text{s}+\text{t}\sqrt{2\text{fs}}+2\text{s}=5\text{s}$
$\Rightarrow \text{t}\sqrt{2\text{fs}}=2\text{s}$
 $\Rightarrow \text{s}=\frac{1}{2}\text{ft}^2$

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