position of a particle moving along the x axis is given by x=2+8t-4t^2 where t is time in seconds. The distance travelled by the particle in first 2 seconds is how much?

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Solution

$P o s i t i o n o f t h e p a r t i c l e i s x = 2 + 8 t - 4 t^{2} a t t = 2 s d i s \tan c e t r a v e l l e d i n 2 s e c o n d i s x = 2 + 8 (2) + 4 (2)^{2} x = 2 + 16 - 16 x = 2 m$

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x = 2 + 8 t - 4 t^{2}

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