During its motion along a $30 k m$ path, a car takes $10 m i n$ to cover first $10 k m$ , $15 m i n$ to cover next $10 k m$ and $20 m i n$ to cover the rest of the distance. So, the body moves with:

constant speed

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acceleration

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retardation

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none of these

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Solution

The correct option is B retardation
Speed of the car during first part $= \frac{10}{10} k m / m i n = 1 k m / m i n$
Speed during second part $= \frac{10}{15} k m / m i n = \frac{2}{3} k m / m i n$
Speed during third part $= \frac{10}{20} k m / m i n = \frac{1}{2} k m / m i n$
So, the speed reduces with time.The car moves with retardation.

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