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

Answer:

20.996 m

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,

$a=\frac{v-u}{t_{Total}}=\frac{12.4062-0}{7.13}=1.74$ m/s²

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

Displacement is given as:

$s=ut+\frac{1}{2} at^{2}$

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

This gives,

$s=0+\frac{1}{2} \times 1.74 \times (4.91257)^{2}=20.996 \textrm{ m}$

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