Merg Alignment 4P A local elevator moves upward at a constant 2.6 mps passing a

Question

Merg Alignment 4P A local elevator moves upward at a constant 2.6 mps passing a stopped express elevator. Precisely 1.3 seconds later the express elevator starts upward with a constant acceleration and catches up with the local elevator where the velocity of the local with 2.60 respect to the express elevator is -9.5 m/s. Determine (a) the acceleration of the express elevator in m/s/s and (b) the distance traveled in m for the express to catch up with the local elevator v lo 5p An attacking projectile A is launched from an elevation of 3.9 m with an initial vlci ht A

Explanation / Answer

4P:

for constant speed elevator,

v1 = 2.6 m/s

after time t, d = v1 t = 2.6 t

Now for other elevator,

t2 = t - 1.3 s

vi = 0 and vf = 9.5 + 2.6 = 12.1 m/s

vf = vi + a t

12.1 = 0 + a(t - 1.3)

a = 12.1 / (t - 1.3)

and d = v0 t + a t^2 /2

d = (12.1 / (t - 1.3))(t - 1.3)^2 /2

d = 6.05(t - 1.3)

so 2.6 t = 6.05(t - 1.3)

t = 2.28 s

(A) a = (12.1) / (2.28 - 1.3)

a = 12.35 m/s^2

(B) d = 2.6 t = 5.93 m

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---