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Answer :
Final answer:
Hence, the car covers a distance of 265.4 meters in 9 seconds.
Explanation:
A car moving with an initial velocity of 35.9 m/s slows down at a constant rate of -1.4 m/s². To find its velocity after 9 seconds of deceleration and the distance it covers in this time, we use the equations of motion.
Velocity after 9 seconds
The final velocity (v) can be calculated using the equation v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values: v = 35.9 m/s + (-1.4 m/s²)(9 s) = 35.9 m/s - 12.6 m/s = 23.3 m/s. Thus, the car's velocity after 9 seconds is 23.3 m/s.
Distance covered in this time
To find the distance (s), we use the equation s = ut + ½at². Therefore, s = (35.9 m/s)(9 s) + ½(-1.4 m/s²)(9 s)² = 322.1 m - 56.7 m = 265.4 m.
Hence, the car covers a distance of 265.4 meters in 9 seconds.
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