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A car is accelerating as it passes two checkpoints that are 35.17 meters apart. The car's speed at the first checkpoint is 5.28 m/s. If the car's acceleration is 1.70 m/s², what is its speed at the second checkpoint?

Answer :

Final answer:

To calculate the car's speed at the second checkpoint given its speed at the first checkpoint, acceleration, and distance covered, we use the kinematic formula for final velocity. The calculation involves the use of the formula v2 = u2 + 2as, where v is the final speed, u is the initial speed, a is the acceleration, and s is the distance traveled.

The speed of the car at the second checkpoint is approximately 12.16 m/s.

Explanation:

To solve this problem, we can use the equations of motion. The equation that relates the final velocity (v), initial velocity (u), acceleration (a), and displacement (s) is v^2 = u^2 + 2as. In this case, the initial velocity is 5.28 m/s, the acceleration is 1.70 m/s², and the displacement is 35.17 meters. Plugging in these values, we can solve for the final velocity:

v^2 = 5.28^2 + 2(1.70)(35.17)

v^2 = 27.98 + 120

v^2 = 147.98

v = √147.98

v ≈ 12.16 m/s

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