KCET · Physics · Laws of Motion
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
- A \( 640 \mathrm{~m} \)
- B \( 320 \mathrm{~m} \)
- C \( 1280 \mathrm{~m} \)
- D \( 160 \mathrm{~m} \)
Answer & Solution
Correct Answer
(D) \( 160 \mathrm{~m} \)
Step-by-step Solution
Detailed explanation
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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