Answer :
Answer with Explanation:
We are given that
Mass of superman=m=78 kg
Mass of train=m'=17863 kg
Speed of train=u'=75 km/h=[tex]75\times \frac{5}{18}=20.8 m/s[/tex]
[tex]1 km/h=\frac{5}{18} m/s[/tex]
Let initial speed of superman=u
Momentum=mv
Using the formula
[tex]78u=17863\times 75[/tex]
[tex]u=\frac{17863\times 75}{78}[/tex]
[tex]u=17175.9 km/h[/tex]
Average horizontal force=0.58
Deceleration [tex]a=-0.58\times 9.8=-5.68 m/s^2[/tex]
Final speed of train=v'=0
[tex]v=u+at[/tex]
Using the formula
[tex]0=20.8-5.68t[/tex]
[tex]5.68t=20.8[/tex]
[tex]t=\frac{v'-u'}{a}=\frac{0-20.8}{-5.68}[/tex]
[tex]t=3.66 s[/tex]
[tex]v^2-u^2=2as[/tex]
Using the formula
[tex]0-(20.8)^2=2(-5.68)a[/tex]
[tex](20.8)^2=2(5.68)s[/tex]
[tex]s={\frac{(20.8)^2}{2(5.68)}=38.1 m[/tex]
Answer:
Explanation:
mass of superman, m = 78 kg
mass of train, M = 17863 kg
(a)
initial velocity of train, U = 75 km/h = 20.83 m/s
let u is the velocity of superman.
By using the conservation of momentum
m x u = M x V
78 x u = 17863 x 75
u = 17175.96 km/h = 4771.1 m/s
(b)
a = - 0.58 x 9.8 = - 5.684 m/s²
by first equation of motion
v = U + at
0 = 20.83 - 5.684 t
t = 3.66 s
use second equation of motion
s = Ut + 0.5 x at²
s = 20.83 x 3.66 - 0.5 x 5.684 x 3.66 x 3.66
s = 76.24 - 38.1
s = 38.1 m