Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac
h bus drives a constant 55 mph, find the rate at which the distance between the buses is changing when they are 13mi apart, heading toward each other.
1 answer:
Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.
Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
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Answer:
the train is moving at the speed of v = 1.79 m/s
Explanation:
given,
rain drop is falling vertically down with the speed of = 3.84 m/s
angle of the rain drop = 25°
tan θ =
tan 25° =
v =3.84 × tan 25°
v = 1.79 m/s
hence, the train is moving at the speed of v = 1.79 m/s
Answer:
a) θ = 58.3º
b) vfh = 13.7 m/s
c) g = -9.8 m/s2
d) h = 22.2 m
e) vfb = 15.5 m/s
Explanation:
a)
- Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
- Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
- vₓ₀ = v * cos θ (1)
- where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
- Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:
- Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:
- ⇒θ = cos⁻¹ (0.526) = 58.3º (4)
b)
- At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
- Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
- vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)
c)
- At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)
d)
- Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:
- Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
- v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
- Replacing (7) in (6), we get:
e)
- When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
- The horizontal component, since it keeps constant, is just v₀x:
- v₀ₓ = 13.7 m/s
- The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:
- Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:
- Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:
Let be the average acceleration over the first 2.46 seconds, and
the average acceleration over the next 6.79 seconds.
At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let be the velocity of the car after the first 2.46 seconds.
By definition of average acceleration, we have
and we're also told that
(or possibly the other way around; I'll consider that case later). We can solve for in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:
Now we can solve for . We find that
In the case that the ratio of accelerations is actually
we would instead have
in which case we would get a velocity of