The motion of a particle is described by the equation x = at + $b t^{2}$ , where a = 15 cm $s^{- 1}$ and b = 3 cm $s^{- 2}$ . Its instantanous velocity at t = 3 s will be

33 cm

s^{- 1}

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18 cm

s^{- 1}

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16 cm

s^{- 1}

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32 cm

s^{- 1}

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Solution

The correct option is D 33 cm $s^{- 1}$
Given:
$a = 15 c m s^{- 1} b = 3 c m s^{- 2}$
$x = a t + b t^{2}$
$\frac{d x}{d t} = v = a + 2 b t ∣_{t = 3}$
$= 15 + 2 \times 3 \times 3$
$= 15 + 18$
$V = 33 c m / s$

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The motion of a particle is described by the equation x = at + bt2, where a = 15 cms−1 and b = 3 cms−2. Its instantanous velocity at t = 3 s will be

The correct option is D 33 cm s−1Given:a=15cms−1b=3cms−2x=at+bt2dxdt=v=a+2bt∣t=3 =15+2×3×3 =15+18 V=33cm/s

The motion of a particle is described by the equation x = at + $b t^{2}$ , where a = 15 cm $s^{- 1}$ and b = 3 cm $s^{- 2}$ . Its instantanous velocity at t = 3 s will be

The correct option is D 33 cm $s^{- 1}$
Given:
$a = 15 c m s^{- 1} b = 3 c m s^{- 2}$
$x = a t + b t^{2}$
$\frac{d x}{d t} = v = a + 2 b t ∣_{t = 3}$
$= 15 + 2 \times 3 \times 3$
$= 15 + 18$
$V = 33 c m / s$