Question

015 10.0 points

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.

Answer 1

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.