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 + 2asv2=u2+2as
Where:
- vvv is the final velocity (0 m/s, as the car comes to rest)
- uuu is the initial velocity (40 m/s)
- aaa is the acceleration (unknown)
- sss is the distance covered (unknown)
Given that v=0v = 0v=0, u=40 m/su = 40 , ext{m/s}u=40m/s, and the time t=20 st = 20 , ext{s}t=20s, we can solve for sss.
First, let's rearrange the equation:
0=(40)2+2as0 = (40)^2 + 2as0=(40)2+2as
Simplifying further:
0=1600+2as0 = 1600 + 2as0=1600+2as
Since the car comes to rest, its final velocity is 0 m/s. We can now solve for sss:
−1600=2as-1600 = 2as−1600=2as
Dividing both sides by 2a:
−800=as-800 = as−800=as
Now, substituting the time <span