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Q.

A particle starts rest and has an acceleration of 2m/s² for 10 sec. After that, it travels for 30 sec with constant speed and then undergoes a retardation of 4m/s²and comes back to rest. The total distance covered by the particle is

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a

800 m

b

700 m

c

750 m

d

650 m

answer is B.

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Detailed Solution

$S_{1} = \frac{1}{2} {at}_{1}^{2} = \frac{1}{2} \times 2 \times 100 = 100 m$
Question Image
S₂ = vt₂ = (at₁)t₂ = 2 x 10 x 30 = 600m
$\begin{array}{l} v^{2} - u^{2} = 2 {as}_{3} \Rightarrow 0^{0} - (20)^{2} = - 2 \times 4 \times S_{3} \\ S_{3} = \frac{- 400}{- 8} = 50 m \end{array}$
S = S₁ + S₂ + S₃ = 750m

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