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A local train, travelling at 72 km/hr is brought to rest in 10 seconds by applying the brake. How much is the acceleration produced in this case? Also how much is distance (s) covered by the train before coming to rest?

A. a=2 m/s²,s=300m
B. a=−2 m/s²,s=300m
C. a=2 m/s²,s=100m
D. a=−2 m/s²,s=100m

1 Answer

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Final answer:

The acceleration produced by the brakes is -2 m/s² and the distance covered by the train before coming to rest is 100 meters. The correct answer is option C.

Step-by-step explanation:

In order to determine the acceleration produced by the brakes, we need to convert the speed from kilometers per hour to meters per second.

The conversion factor is 1000 meters = 1 kilometer and 1 hour = 3600 seconds.

So, the speed of the train in meters per second is

72 km/hr * (1000 m/1 km) * (1 hr/3600 s) = 20 m/s.

Next, we can use the formula for acceleration, which is acceleration (a) = change in velocity (Δv) / change in time (Δt).

In this case, the change in velocity is 20 m/s - 0 m/s = 20 m/s, and the change in time is 10 seconds.

So, the acceleration is 20 m/s / 10 s = 2 m/s².

To find the distance covered by the train before coming to rest, we can use the formula for distance, which is

distance (s) = initial velocity (u) * time (t) + 0.5 * acceleration (a) * time^2.

In this case, the initial velocity is 20 m/s, the time is 10 seconds, and the acceleration is -2 m/s² (negative because the train is slowing down).

Plugging in these values, we get:

distance (s) = 20 m/s * 10 s + 0.5 * (-2 m/s²) * (10 s)^2

= 200 m - 100 m = 100 m.

Therefore, the correct answer is option D. a=−2 m/s², s=100m.

User Manish Dubey
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