The driver of a car moving at a speed of 10 m/s applies breaks after noticing an obstacle. Retardation due to the breaks is $4 m / s^{2}$ . Distance travelled by the car in 3 seconds after applying breaks is

12 m

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12.5 m

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10 m

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11 m

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Solution

The correct option is A 12 m
$S = u t + \frac{1}{2} a t^{2} = (10 \times 3) - \frac{1}{2} \times 4 \times 3^{2}$
$\Rightarrow S = 12 m$

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Similar questions

Q. The driver of a car moving at a speed of 10 m/s applies breaks after noticing an obstacle. Retardation due to the breaks is

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A car moving with a speed of $40 k m / h r$ can be stopped by applying breaks for $2$ km. If the same car is moving with a speed of $80 k m / h r$ , what is the minimum stopping distance? Assume constant retardation.

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The driver of a car moving at a speed of 10 m/s applies breaks after noticing an obstacle. Retardation due to the breaks is 4 m/s2. Distance travelled by the car in 3 seconds after applying breaks is

The correct option is A 12 mS=ut+12at2=(10×3)−12×4×32 ⇒S=12 m

The driver of a car moving at a speed of 10 m/s applies breaks after noticing an obstacle. Retardation due to the breaks is $4 m / s^{2}$ . Distance travelled by the car in 3 seconds after applying breaks is

The correct option is A 12 m
$S = u t + \frac{1}{2} a t^{2} = (10 \times 3) - \frac{1}{2} \times 4 \times 3^{2}$
$\Rightarrow S = 12 m$