Question

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:

Displacement = 20.996 m

Explanation: v = 12.4062 m/s ; t = 7.13 s

v= u + at

Here u = 0 ( Since it starts from rest)

  substituting the values

12.4062 = 0 + a × 7.13

Therefore a = $\frac{12.4062}{7.13}$

                    = 1.74

Now use,

s = ut +$\frac{1}{2}$ at²

Where u = 0

s = 0 + $\frac{1}{2}$ × 1.74 × 4.91257 × 4.91257

 = 20.996 m