You throw a balloon that floats in the air with a velocity of 2 m / s south . If the wind speed is 5 m / s west , how far south
1 answer:
Answer:
The distance traveled by the balloon is 10.77 m
Explanation:
velocity of the ball, = 2 m/s south
velocity of the air, = 5 m/s west
To determine the distance the balloon will travel after 2 seconds, first determine the resultant velocity of the balloon.
| 2m/s
|
|
↓
5m/s ←------------------
the two velocities forms a right angled triangle and the resultant will be the hypotenuses side of the triangle.
R² = 5² + 2²
R² = 29
R = √29
R = 5.385 m/s
The distance traveled by the balloon is calculated as;
d = R x t
where;
t is time of the motion = 2 seconds
d = 5.385 x 2
d = 10.77 m
Therefore, the distance traveled by the balloon is 10.77 m.
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Answer:
(a) 1.939 m/h
(b) 0.926 m/h
(c) -0.315 m/h
(d) -1.21 m/h
Explanation:
Here, we have the water depth given by the function of time;
D(t) = 7 + 5·cos[0.503(t-6.75)]
Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;
D'(t) =
= 5×(-sin(0.503(t-6.75))×0.503
= -2.515×(-sin(0.503(t-6.75))
= -2.515×(-sin(0.503×t-3.395))
Therefore we have;
(a) at 5:00 AM = 5 - 0:00 = 5
D'(5) = -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h
(b) at 6:00 AM = 6 - 0:00 = 6
D'(5) = -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h
(c) at 7:00 AM = 7 - 0:00 = 7
D'(5) = -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h
(d) at Noon 12:00 PM = 12 - 0:00 = 12
D'(5) = -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.