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sleet_krkn [62]

1 year ago

6

A girl is running toward the front of a train at 10 m/s. If the train is going 75 m/s on the Southbound tracks, what is the spee

d of the train relative to the girl?

2 answers:

sineoko [7]1 year ago

6 0

Answer:

velocity of the train with respect to girl is 85 m/s

Explanation:

As we know that girl is moving towards the train in front of it with speed

$v_1 = 10 m/s$

also we know that the speed of the train towards the girl is

$v_2 = 75 m/s$

Now we know that relative velocity of the train with respect to girl is given as

$\vec v_{tg} = \vec v_t - \vec v_g$

here let say the velocity of train is positive while girl is running opposite to train so its velocity is negative

$\vec v_{tg} = 75 m/s - (-10 m/s)$

$\vec v_{tg} = 85 m/s$

dimulka [17.4K]1 year ago

5 0

We can take as positive direction the direction of the train, and as negative direction the direction of the girl. The speed of the train relative to the girl will be given by the difference between the two velocities:
$v' = v_t - v_g$
which means
$v'=75 m/s - (-10 m/s)=85 m/s$

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Answer:

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c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

$x*g = x * \frac{dv}{dt} + v^2$

- Where, v = Speed of the raindrop

- Proposed solution to given ODE:

v = a*t

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Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

a = dv / dt ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

v = dx / dt , v = a*t ... proposed solution (Eq 2)

v .dt = dx = a*t . dt

- integrate both sides:

∫a*t . dt = ∫dt

x = 0.5*a*t^2 ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

= 1.5 a^2 t^2

a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

x (t) = 0.5*a*t^2

x (t = 3.0) = 0.5*9.81*3^2 / 3

x (3.0) = 14.7 m

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

m (t) = k*x (t)

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Explanation:

It is given that,

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$d_x=v_x\times t\\\\=3.6\times 2\\\\=7.2\ m$

Let $d_y$ is the height of the cliff. It can be calculated using second equation of motion as follows :

$d_y=u_yt+\dfrac{1}{2}gt^2\\\\d_y=\dfrac{1}{2}\times 9.8\times 2^2\\\\=19.6\ m$

So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.

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