A racing car accelerates on a straight road from rest to a speed of 50

A racing car accelerates on a straight road from rest to a speed of 50 m/s in 25 s. Assuming uniform acceleration of the car throughout, the distance covered in this time will be

625 m
1250 m
2500 m
50 m
This question was previously asked in
UPSC NDA-1 – 2016
The distance covered by the racing car is 625 m.
We are given the initial velocity (u = 0 m/s, since it starts from rest), final velocity (v = 50 m/s), and time (t = 25 s). Assuming uniform acceleration (a), we can use the kinematic equations. First, find the acceleration using v = u + at: 50 = 0 + a * 25, which gives a = 50/25 = 2 m/s². Then, find the distance (s) using s = ut + (1/2)at²: s = (0 * 25) + (1/2) * 2 * (25)² = 0 + 1 * 625 = 625 m. Alternatively, the average velocity is (u+v)/2 = (0+50)/2 = 25 m/s. Distance = Average velocity * time = 25 m/s * 25 s = 625 m.
This problem involves basic kinematics under constant acceleration. The relevant equations of motion are v = u + at, s = ut + (1/2)at², and v² = u² + 2as. Any of these can be used depending on the given and required variables.
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