At a distance of $500 m$ from the traffic light,brakes are applied to an automobile moving at a velocity of $20 m / s$ .The position of the automobile relative to the traffic light $50 s$ after applying brakes ,if its acceleration is $- 0.5 m / s^{2}$ is

125 m

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375 m

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400 m

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100 m

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Solution

The correct option is C $125 m$
First find that time in which the vehicle stops.
From $\to v = \to u + \to a t \Rightarrow 0 = 20 - 0.5 t \Rightarrow t = 40 s$ .
The distance covered by the vehicle in $40 s$ would be $s = 20 \times 40 - \frac{1}{2} \times 0.5 \times (40)^{2}$
$= 800 - 400 = 400 m$
Required distance $= 500 - 400 = 100 m$

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At a distance of 500m from the traffic light,brakes are applied to an automobile moving at a velocity of 20m/s.The position of the automobile relative to the traffic light 50s after applying brakes ,if its acceleration is −0.5m/s2 is

The correct option is C 125mFirst find that time in which the vehicle stops.From →v=→u+→at ⇒0=20−0.5 t ⇒t=40 s. The distance covered by the vehicle in 40 s would be s=20×40−12×0.5×(40)2=800−400=400 mRequired distance =500−400=100 m

At a distance of $500 m$ from the traffic light,brakes are applied to an automobile moving at a velocity of $20 m / s$ .The position of the automobile relative to the traffic light $50 s$ after applying brakes ,if its acceleration is $- 0.5 m / s^{2}$ is