Answer:
1534100 N in the opposite direction of his initial velocity
Explanation:
1 cm = 0.01 m
So the person is subjected to a deceleration that stops his initial motion of 23 m/s within a distance of 0.01 m. We can use the following equation of motion to find out the deceleration:
[tex]v^2 - v_0^2 = 2a\Delta s[/tex]
where v = 0 m/s is the final velocity of the person, [tex]v_0[/tex] = 23m/s is the initial velocity of the person when hit, a is the deceleration, and [tex]\Delta s = 0.01 m[/tex] is the distance traveled:
[tex]0^2 - 23^2 = 2*a*0.01[/tex]
[tex]a = -23^2 / (2*0.01) = -26450 m/s^2[/tex]
We can calculate the average force he's subjected to using Newton's2nd law:
[tex]F = am = -26450 * 58 = -1534100 N[/tex]
So he is subjected to a force of 1534100 N in the opposite direction of his initial velocity.
