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stich3 [128]
3 years ago
12

1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc

al train 100 m ahead on the same track and traveling in the same direction at a constant 11 m/s. The local engineer is unaware of the situation. The express engineer jams on the brakes and slows the express at 3.0 m/s2. (a) To determine whether the trains collide, use kinematics to calculate their positions when the express train stops: i. How much time will it take the express train to stop? ii. In that time, how far will the express train have moved? iii. In that time, how far will the local train have moved? iv. Based on the results of ii. and iii., do the trains collide? Explain.
Physics
1 answer:
Colt1911 [192]3 years ago
6 0

Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of 3.0 m/s^2.

So, the acceleration of the express train, a=-3 m/s^2.

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, a=-3 m/s^2

So, 0=36+(-3)t

\Rightarrow t= 36/3=12 seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

v^2=u^2+2as

\Rightarrow 0^2=36^2+2(-3)s

\Rightarrow s=\frac{36\times36}{6}

\Rightarrow s=216 meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m  [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

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Answer:

The time taken by the bar to reach the bottom t=4.886s

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To find:

How long it takes to reach the bottom ‘t’

<u>Step by Step Explanation:</u>

Solution:

We know that the formula for weight of the soap bar is given as

F_{g}=m g \sin \theta

The frictional force acting on this soap bar is determined by

F_{f}=\mu m g \cos \theta

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F=m a

Here F=F_{g}-F_{f} and thus

F_{g}-F_{f}=m am g \sin \theta-\mu m g \cos \theta=m a

a=g \sin \theta-\mu g \cos \theta

WhereF_{g}=Force imparted due to weight

F_{f}=Frictional Force

m=Mass of the bar

g=Acceleration due to gravity

a=Acceleration of the bar

\sin \theta and \cos \theta are the angles involved in the system

If the bar starts from the rest

Equations of motion involved in calculating the displacement of the bar is given as

s=\frac{1}{2} a t^{2}, From this

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t^{2}=\frac{2 s}{a}

t=\sqrt{\frac{2 s}{a}}

Where s= displacement or length moved by the bar

a=Acceleration of the bar

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Substitute all the known values in the above equation we get

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a=g \sin \theta-\mu g \cos \theta

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t=\sqrt{23.87332}

t=4.886s

Result:

Thus the time taken by the bar to reach the bottom is t=4.886s

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