An electron starting from rest has a velocity that increases linearly with time as $v = k t$ , where $k = 2 m s^{- 2}$ . The distance travelled in the first 3 seconds will be

9 m

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16 m

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27 m

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36 m

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Solution

The correct option is A
9 m

Integrating velocity w.r.t time to obtain distance, we have

$S = \int_{0}^{3} v d t = \int_{0}^{3} k t d t = {[\frac{1}{2} k t^{2}]}_{0}^{2} = \frac{1}{2} \times 2 \times 9 = 9 m$

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An electron starting from rest has a velocity that increases linearly with time as v=k t, where k=2ms−2. The distance travelled in the first 3 seconds will be

The correct option is A 9 m Integrating velocity w.r.t time to obtain distance, we have S=∫30vdt=∫30ktdt=[12kt2]30=12×2×9=9m

An electron starting from rest has a velocity that increases linearly with time as $v = k t$ , where $k = 2 m s^{- 2}$ . The distance travelled in the first 3 seconds will be

The correct option is A
9 m

Integrating velocity w.r.t time to obtain distance, we have

$S = \int_{0}^{3} v d t = \int_{0}^{3} k t d t = {[\frac{1}{2} k t^{2}]}_{0}^{2} = \frac{1}{2} \times 2 \times 9 = 9 m$