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oksian1 [2.3K]
3 years ago
7

The engine of a locomotive exerts a constant force of 8.1*10^5 N to accelerate a train to 68 km/h. Determine the time (in min) t

aken for the train of mass 1.9*10^7 kg to reach this speed from rest.
Physics
2 answers:
Bumek [7]3 years ago
8 0

Answer:443.1 s

Explanation:

Given

Engine of a locomotive exerts a force of 8.1\times 10^5 N

Mass of train=1.9\times 10^7

Final speed (v)=68 km/h \approx 18.88 m/s

F=ma

so acceleration(a) =\frac{F}{m}=\frac{8.1\times 10^5}{1.9\times 10^7}

a=0.042631 m/s^2

and acceleration is

a=\frac{v-u}{t}

0.042631=\frac{18.88-0}{t}

t=443.089 \approx 443.1 s

nalin [4]3 years ago
5 0

Explanation:

It is given that,

Force acting on the engine, F=8.1\times 10^5\ N

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, m=1.9\times 10^7\ kg

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 :

a=\dfrac{F}{m}

a=\dfrac{8.1\times 10^5}{1.9\times 10^7}

a=0.042\ m/s^2

Now using first equation of motion to find time taken as :

v=u+at

t=\dfrac{v-u}{a}

t=\dfrac{18.88-0}{0.042}

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.

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