Ms. Reitman's scooter starts from rest and the final velocity is
1 answer:
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Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v =
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
t =
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m
Answer:
(A) 88.92 cm
(B) 22.22 cm
Explanation:
distance (s) = 200 cm = 0.2 m
initial velocity (v) = 0 m/s
acceleration due to gravity (g) = 9.8 m/s^{2}
lets first find the time (T) it takes for the first drop to strike the floor
from s = ut +
200 = 0 +
200 =
200 / 4.9 =
T = 6.4
(A) When the first drop strikes the floor, how far below the nozzle is the second drop.
we can find how far the second drop was when the first drop hits the ground from the formula s = ut +
where
- s = distance
- u = initial velocity = 0
- t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the second drop = 2/3 of the time it takes the first drop to strike the ground (
)
- a = acceleration due to gravity = 9.8 m/s^{2}
substituting all required values we have
s = 0 + ()
s = 0 + ()
s = 88.92 cm
(B) When the first drop strikes the floor, how far below the nozzle is the third drop.
we can find how far the third drop was when the first drop hits the ground from the formula s = ut +
where
- s = distance
- u = initial velocity = 0
- t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the third drop = 1/3 of the time it takes the first drop to strike the ground (
- a = acceleration due to gravity = 9.8 m/s^{2}
substituting all required values we have
s = 0 + ()
s = 0 + ()
s = 22.22 cm
