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A car, starting from rest, accelerates at the rate f through a distance  s , then continues at constant speed for time t and then decelerates at the rate f/2  to come to rest. If the total distance traversed is 15 s , then

Option: 1

\frac{ft^2}{72}


Option: 2

s=\frac{1}{4}ft^{2}


Option: 3

s=ft


Option: 4

s=\frac{1}{6}ft^{2}


Answers (1)

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For first part of journey ,s = s1,

s_{1}= \frac{1}{2}f\: t_{1}^{2}= s\cdots \cdots \cdots (i)

\nu = f\: t_{1}\cdots \cdots \cdots (ii)

For second part of journey,

s_{2}= \nu t

or\: \: s_{2}= f\; t_{1}\; t\cdots \cdots \cdots (iii)

For the third part of journey,

s_{3}= \frac{1}{2}\left ( \frac{f}{2} \right )\left ( 2t_{1} \right )^{2}

or\: \: s_{3}= \frac{1}{2}\times \frac{4ft_{2}^{1}}{2}

or\: \: s_{3}=2s_{1}= 2s\cdots \cdots \cdots \left ( iv \right )

or\: \: s_{1}+s_{2}+s_{3}= 15s

s+ft_{1}\; t+2s= 15s

or\: \: ft_{1}\; t= 12s\cdots \cdots \cdots \left ( v \right )

From (i) and (v)

\frac{s}{12s}= \frac{ft_{1}^{2}}{2\times ft_{1}t}

or\: \: t_{1}= \frac{t}{6}

or\: \: s= \frac{1}{2}ft_{1}^{2}= \frac{1}{2}f\left ( \frac{t}{6} \right )^{2}=\frac{ ft^{2}}{72}

 

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Ritika Jonwal

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