as I explained for this may this helps you!
The applicable kinematic expression is:
where v is final velocity, u is initial velocity, a is acceleration and s is displacement.
Since the car needs to stop, hence in the first instance distance traveled , when r is the retardation
in the second instance given
Hence, distance traveled with same deceleration
Fractional increase in stopping distance
∴Percent increase in stopping distance=0.44×100%=44%
depends on the force added to stop the car...
Step-by-step explanation: i dont know the answer but al the answers are complocated
HOPE IT HELPS
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what ? n
i the question is incomplete.