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a jogger accelerates from rest to 3.0m/s in 2.0s A car accelerates from 38.0 to 41.0 m/s also in 2.0 s. Find the acceleration of the jogger and determine the acceleration of the car. Does the car travel further than the jogger during the 2.0 s? If so how much further?

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Final answer:

The jogger and the car both have an acceleration of 1.5 m/s². The car travels further than the jogger during the 2.0-second time interval. The jogger covers a distance of 3.0 meters, while the car covers a distance of 77.0 meters.

Step-by-step explanation:

The acceleration of the jogger can be calculated using the equation:

acceleration = (final velocity - initial velocity) / time

Plugging in the given values, we get:

acceleration = (3.0 m/s - 0) / 2.0 s = 1.5 m/s²

The acceleration of the car can be calculated in the same way:

acceleration = (41.0 m/s - 38.0 m/s) / 2.0 s = 1.5 m/s²

Since both the jogger and the car have the same acceleration, they both cover the same distance during the 2.0-second time interval. The distance covered by an object under constant acceleration can be calculated using the equation:

distance = initial velocity * time + (0.5 * acceleration * time²)

Plugging in the given values for the jogger:

distance = 0 * 2.0 s + (0.5 * 1.5 m/s² * (2.0 s)²) = 3.0 meters

Plugging in the given values for the car:

distance = 38.0 m/s * 2.0 s + (0.5 * 1.5 m/s² * (2.0 s)²) = 77.0 meters

User RJardines
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