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The car started to move along a straight road.

Physics

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D) Car start to move along straight road.

  1. Analyze the change in distance over equal time intervals.

    The car starts from rest (0 m0 \, \text{m} at 0 s0 \, \text{s}). We observe the distance covered in successive 3 s3 \, \text{s} intervals:

    • From 0 s0 \, \text{s} to 3 s3 \, \text{s}: 0.8 m0.8 \, \text{m}
    • From 3 s3 \, \text{s} to 6 s6 \, \text{s}: 2.0−0.8=1.2 m2.0 - 0.8 = 1.2 \, \text{m}
    • From 6 s6 \, \text{s} to 9 s9 \, \text{s}: 4.3−2.0=2.3 m4.3 - 2.0 = 2.3 \, \text{m}
    • From 9 s9 \, \text{s} to 12 s12 \, \text{s}: 8.0−4.3=3.7 m8.0 - 4.3 = 3.7 \, \text{m}
    • From 12 s12 \, \text{s} to 15 s15 \, \text{s}: 11.5−8.0=3.5 m11.5 - 8.0 = 3.5 \, \text{m}
    • From 15 s15 \, \text{s} to 18 s18 \, \text{s}: 14.0−11.5=2.5 m14.0 - 11.5 = 2.5 \, \text{m}
    • From 18 s18 \, \text{s} to 21 s21 \, \text{s}: 14.8−14.0=0.8 m14.8 - 14.0 = 0.8 \, \text{m}
  2. Describe the motion based on the changing distances.

    The distance covered in each 3 s3 \, \text{s} interval initially increases (from 0.8 m0.8 \, \text{m} to 3.7 m3.7 \, \text{m}), indicating that the car is accelerating. After the 12 s12 \, \text{s} mark, the distance covered in each 3 s3 \, \text{s} interval decreases (from 3.5 m3.5 \, \text{m} to 0.8 m0.8 \, \text{m}), indicating that the car is decelerating.

    The motion of the car is:
    The car starts from rest, accelerates for approximately the first 12 s12 \, \text{s} to reach a maximum speed, and then decelerates for the remainder of the observed period.

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