An engine of a vehicle can produce a maximum acceleration of $4 m s^{- 2}$ . Its brakes can produce a maximum retardation of $6 m s^{- 2}$ . The minimum time at which it can cover a distance of $3 k m$ is:

30 s

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40 s

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50 s

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60 s

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Solution

The correct option is C 50 s
maximum acceleration $= 4 m / s^{2}$
maximum reduction $= 6 m / s^{2}$
distance $= 3 k m$
$[\frac{α β}{2 (α + β)}] \times t^{2} = S$
$α =$ maximum acceleration
$β =$ maximum reduction
$t =$ time
$S =$ distance
$t^{2} =$ 2500
$t =$ 50 sec

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An engine of a vehicle can produce a maximum acceleration of 4ms−2. Its brakes can produce a maximum retardation of 6ms−2. The minimum time at which it can cover a distance of 3km is:

The correct option is C 50 smaximum acceleration =4m/s2maximum reduction =6m/s2distance =3km [αβ2(α+β)]×t2=S α= maximum accelerationβ= maximum reduction t= time S= distance t2= 2500 t= 50 sec

An engine of a vehicle can produce a maximum acceleration of $4 m s^{- 2}$ . Its brakes can produce a maximum retardation of $6 m s^{- 2}$ . The minimum time at which it can cover a distance of $3 k m$ is:

The correct option is C 50 s
maximum acceleration $= 4 m / s^{2}$
maximum reduction $= 6 m / s^{2}$
distance $= 3 k m$
$[\frac{α β}{2 (α + β)}] \times t^{2} = S$
$α =$ maximum acceleration
$β =$ maximum reduction
$t =$ time
$S =$ distance
$t^{2} =$ 2500
$t =$ 50 sec