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A subway train starting from rest leaves a
station with a constant acceleration. At the
end of 7.13 s, it is moving at 12.4062 m/s.
What is the train's displacement in the first
4.91257 s of motion?
Answer in units of m.

User HanSooloo
by
8.6k points

1 Answer

3 votes

Answer:

20.996 m

Step-by-step explanation:

Given:

Initial velocity,
u=0 \textrm{ m/s}

Final velocity,
v=12.4062 \textrm{ m/s}

Total time taken,
t_(Total) = 7.13 s.

∴ Acceleration is given as,


a=(v-u)/(t_(Total))=(12.4062-0)/(7.13)=1.74 m/s²

Now, using Newton's equation of motion, we find the displacement.

Displacement is given as:


s=ut+(1)/(2) at^(2)

Plug in 0 for
u, 4.91257 for
t and 1.74 for
a. Solve for
s.

This gives,


s=0+(1)/(2) * 1.74 * (4.91257)^(2)=20.996 \textrm{ m}

Therefore, the train's displacement in the first 4.91257 s of motion is 20.996 m.

User Sam Su
by
8.4k points