Answer:
x=31.09m
Explanation:
p1=p2
The momentum of flatcar and the momentum of the worker so
The velocity of the worker is:
[tex]m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}[/tex]
The total motion has a total velocity and is
[tex]Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}[/tex]
The time the worker take walking is
[tex]t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s[/tex]
Now the total time and the total velocity determinate the motion of tha flatcar how far has moved
[tex]x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m[/tex]
The time it takes for him to reach the other end of the flatcar is 8.9s.
The distance traveled by the flatcar during this time is 26.7 m.
The given parameters;
Apply the principle of conservation of linear momentum, assuming that their momentum before collision is equal to the momentum after collision.
[tex]m_1u_1 + m_2 u_2 = v(m_1+ m_2)\\\\(250\times 3) + (61 \times 2) = v(250 + 61)\\\\872 = 311v\\\\v = \frac{872}{311} \\\\v = 2.8 \ m/s[/tex]
The time it takes for him to reach the other end of the flatcar is calculated as follows;
[tex]t = \frac{25 \ m}{2.8 \ m/s} \\\\t = 8.9 \ s[/tex]
The distance traveled by the flatcar during this time is calculated as;
d = 3 x 8.9
d = 26.7 m
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