Respuesta :
Answer:443.1 s
Explanation:
Given
Engine of a locomotive exerts a force of [tex]8.1\times 10^5 N[/tex]
Mass of train[tex]=1.9\times 10^7[/tex]
Final speed (v)[tex]=68 km/h \approx 18.88 m/s[/tex]
F=ma
so [tex]acceleration(a) =\frac{F}{m}=\frac{8.1\times 10^5}{1.9\times 10^7}[/tex]
[tex]a=0.042631 m/s^2[/tex]
and acceleration is
[tex]a=\frac{v-u}{t}[/tex]
[tex]0.042631=\frac{18.88-0}{t}[/tex]
[tex]t=443.089 \approx 443.1 s[/tex]
Explanation:
It is given that,
Force acting on the engine, [tex]F=8.1\times 10^5\ N[/tex]
Initial speed of the engine, u = 0 (at rest)
Final speed of the engine, v = 68 km/h = 18.88 m/s
Mass of the train, [tex]m=1.9\times 10^7\ kg[/tex]
We need to find the time taken by the train. Firstly, we will find the acceleration of the engine from Newton's second law of motion as :
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{8.1\times 10^5}{1.9\times 10^7}[/tex]
[tex]a=0.042\ m/s^2[/tex]
Now using first equation of motion to find time taken as :
[tex]v=u+at[/tex]
[tex]t=\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{18.88-0}{0.042}[/tex]
t = 449.52 seconds
or
t = 7.49 minutes
So, the time taken for the train to reach this speed is 7.49 minutes. Hence, this is the required solution.
