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To find the distance covered by the car before it comes to a stop, we can use the equation of motion:

v2=u2+2asv^2 = u^2 + 2as

Where:

  • vv is the final velocity (0 m/s, as the car comes to rest)
  • uu is the initial velocity (40 m/s)
  • aa is the acceleration (unknown)
  • ss is the distance covered (unknown)

Given that v=0v = 0, u=40 m/su = 40 , ext{m/s}, and the time t=20 st = 20 , ext{s}, we can solve for ss.

First, let's rearrange the equation:

0=(40)2+2as0 = (40)^2 + 2as

Simplifying further:

0=1600+2as0 = 1600 + 2as

Since the car comes to rest, its final velocity is 0 m/s. We can now solve for ss:

−1600=2as-1600 = 2as

Dividing both sides by 2a:

−800=as-800 = as

Now, substituting the time <span

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