Answer:
[tex]t=273s[/tex]or [tex]4.56 minutes[/tex]
Explanation:
The force pull of 6.9x10^5 N increase the speed of the train from the rest to 68 km/h
so:
[tex]F=m*a[/tex]
[tex]a=\frac{F}{m}=\frac{6.9x10^5N}{1.0x10^7kg}[/tex]
[tex]a=69x10^{-3}m/s^2[/tex]
Now knowing the acceleration and using the equation of uniform accelerated motion can find the time:
[tex]v_f=v_i+a*t[/tex]
Solve to t:
[tex]v_f=68\frac{km}{h}*\frac{1000m}{km}*\frac{1h}{3600s}=18.8m/s[/tex]
[tex]t=\frac{v_f-v_i}{a}=\frac{18.8m/s-0}{69x10^{-3}m/s^2}[/tex]
[tex]t=273s[/tex]or [tex]4.56 minutes[/tex]
To solve the problem we must know about the concept of Force.
Force is the product of mass and acceleration. therefore, it is given as,
F = m x a
Where F is the force, m is the mass and a is the acceleration of the body.
The time is taken by the train of mass 1.0 ✕ 107 kg is 273.752 seconds.
Given to us
Mass of the train, m = 1.0 ✕ 10⁷ kg
Force needed, F = 6.9 ✕ 10⁵
itial Velocity, u = 0 km/h
Final Velocity, v = 68 km/h = 18.8889 m/s
To find the acceleration of the train we will use the formula of force,
F = m x a
6.9 ✕ 10⁵ = 1.0 ✕ 10⁷ x a
a = 0.069 m/s²
Now, we will use the first equation of motion to find the time taken by the train to increase the speed of the train from the rest to 18.8889 m/s
[tex]v-u= at\\\\18.8889 - 0 = 0.069 \times t\\\\t = 273.752\rm\ sec[/tex]
Hence, the time is taken by the train of mass 1.0 ✕ 10⁷ kg is 273.752 seconds.
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