The distances travelled by a body starting from rest and travelling with uniform acceleration in successive…
The distances travelled by a body starting from rest and travelling with uniform acceleration in successive intervals of time each of one second will be in the ratio:
\(1: 2: 3\)
\(1: 2: 4\)
\(1: 3: 5\)
\(1: 5: 9\)
Solution
Distance traveled (s) is given by:
\(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\)
Now, interval wise distance traveled are:
\(\mathrm{s}_{1}-\mathrm{s}_{0}=\frac{\mathrm{a}}{2}(1)^{2}=\frac{\mathrm{a}}{2}\)
\(s_{2}-s_{1}=\frac{a}{2}(2)^{2}-\frac{a}{2}(1)^{2}=\frac{3 a}{2}\)
\(\mathrm{s}_{3}-\mathrm{s}_{2}=\frac{\mathrm{a}}{2}(3)^{2}-\frac{\mathrm{a}}{2}(2)^{2}=\frac{5 \mathrm{a}}{2}\)
Hence, its in the ratio \(=\frac{\mathrm{a}}{2}: \frac{3 \mathrm{a}}{2}: \frac{5 \mathrm{a}}{2}=1: 3: 5\)
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Asked in: JEE Mains - Motion In One Dimension - Test 2