The equation to be used is for the rectilinear motion at constant acceleration:
x = v₀t + 0.5at²
a = (v-v₀)/t
where
x is distance
v and v₀ is the final and initial velocity
t is time
a is acceleration
Because the acceleration is decelerating, that would be reported as -7.5 m/s². Substituting,
-7.5 = (0 - v₀)/t
v₀ = 7.5 t --> eqn 1
x = v₀t + 0.5at²
60 = (7.5t)(t) + 0.5(-7.5)(t²)
Solving for t,
t = 4s
Thus,
v₀ = 7.5 m/s² * 4s
v₀ = 30 m/s
Answer: λ (wavelength) = 3.3 m → ΔS = 3.3 m
v (speed) = 5.6 m/s → ΔV = 5.6 m/s
T (period) → ΔT = ?
f (frequency) = ?
If:
Now: (if, ΔT = T), Using the formula of the period of a wave, find the frequency:
Explanation: Your Welcome u.u
Sure.
Can I use your answer to part-'a' ?
If the angular acceleration is actually 32 rev/min², than
after 1.2 min, it has reached the speed of
(32 rev/min²) x (1.2 min) = 38.4 rev/min .
Check:
If the initial speed is zero and the final speed is 38.4 rpm,
then the average speed during the acceleration period is
(1/2) (0 + 38.4) = 19.2 rpm average
At an average speed of 19.2 rpm for 1.2 min,
it covers
(19.2 rev/min) x (1.2 min) = 23.04 revs .
That's pretty close to the "23" in the question, so I think that
everything here is in order.
