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A person of 50 kg weight drove a car weighing 950 kg from rest with uniform acceleration for 10s. He then maintained a constant speed for 10 mins. The brakes were then applied, and the car came to rest in 5s. If 2s after the start the speed of the car became 4m/s. What was the total distance traveled by the car?

A. 500 m

B. 750 m

C. 1000 m

D. 1250 m

1 Answer

3 votes

Final answer:

The total distance traveled by the car is 2414 m.

The correct answer is none of all.

Step-by-step explanation:

First, let's calculate the acceleration of the car during the first phase of motion. We can use the formula: acceleration = (final velocity - initial velocity) / time. In this case, the car's initial velocity is 0 m/s and the final velocity is 4 m/s after 2 seconds. So, the acceleration is (4 m/s - 0 m/s) / 2 s = 2 m/s2.

Next, we can calculate the distance traveled during the first phase using the equation: distance = (initial velocity * time) + (0.5 * acceleration * time2). Plugging in the values, we get distance = (0 m/s * 2 s) + (0.5 * 2 m/s2 * (2 s)2) = 4 m.

During the second phase, the car maintains a constant speed for 10 minutes, which is equivalent to 600 seconds. Therefore, the distance traveled during this phase is speed * time = 4 m/s * 600 s = 2400 m.

Lastly, during the third phase, the car decelerates and comes to a stop in 5 seconds. Using the equation: distance = (initial velocity * time) + (0.5 * acceleration * time2), where the initial velocity is 4 m/s (from the previous phase) and the time is 5 s, we get distance = (4 m/s * 5 s) + (0.5 * (-2 m/s2) * (5 s)2) = 10 m.

Therefore, the total distance traveled by the car is 4 m + 2400 m + 10 m = 2414 m. The correct option is A. 500 m

User Adam Soffer
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