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A car moving with a velocity of \( 20 \mathrm{~m} \mathrm{~s}^{-1} \) is stopped in a distance of \( 40 \mathrm{~m} \). If the same car is
travelling at double the velocity, the distance travelled by it for same retardation is
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travelling at double the velocity, the distance travelled by it for same retardation is
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Verified Answer
The correct answer is:
\( 160 \mathrm{~m} \)
Given, velocity of car $=20 \mathrm{~ms}^{-1} ;$ distance $=40 \mathrm{~m}$
We know, distance travelled $=\frac{(\text { velocity })^{2}}{2 \times \text { acceleration }}$
$\Rightarrow$ acceleration
$=\frac{(\text { velocity })^{2}}{\text { distance travelled }}=\frac{(20)^{2}}{40}=\frac{400}{40}=10 \mathrm{~ms}^{-2}$
Therefore, acceleration $=10 \mathrm{~ms}^{-2}$
When Velocity $=2 \times 20=40 \mathrm{~ms}^{-1}$
then, distance travelled $=\frac{(40)^{2}}{10}=\frac{1600}{10}=160 \mathrm{~m}$
Thus, distance travelled by car $=160 \mathrm{~m}$
We know, distance travelled $=\frac{(\text { velocity })^{2}}{2 \times \text { acceleration }}$
$\Rightarrow$ acceleration
$=\frac{(\text { velocity })^{2}}{\text { distance travelled }}=\frac{(20)^{2}}{40}=\frac{400}{40}=10 \mathrm{~ms}^{-2}$
Therefore, acceleration $=10 \mathrm{~ms}^{-2}$
When Velocity $=2 \times 20=40 \mathrm{~ms}^{-1}$
then, distance travelled $=\frac{(40)^{2}}{10}=\frac{1600}{10}=160 \mathrm{~m}$
Thus, distance travelled by car $=160 \mathrm{~m}$
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