Question

What would be the answer and give the steps which led to your answer.

Answer 1

Using 2nd equation of uniformly accelerated motion.

In 10 seconds, distance covered is,
$x1 = ut + \frac{1}{2} a {t}^{2} \\ x2 = \frac{1}{2} \times a \times {10}^{2} \\ x1 = 50a$
in next 20 seconds, the distance covered will be (distance covered in 20 seconds-distance covered in 10 seconds)

$x2 = ut + \frac{1}{2} a {t}^{2} - (ut + \frac{1}{2} a {t}^{2} ) \\ x2 = \frac{1}{2} a {t}^{2} - \frac{1}{2} a {t}^{2} \\ x2 = \frac{1}{2} \times a \times 400 - \frac{1}{2} \times a \times 100 \\ x2 = 200a - 50a \\ x2 = 150a$
50a : 150a
1 : 3

Answer is (c)

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