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## PHY101 ASSIGNMENT 1 SOLUTION FALL 2021

Provide by VU Answer

Due Date: 10 Dec 2021

Total Marks: 20

Question 1:

The driver of a 2.0 × 103 kg red car traveling on the highway at 45m/s slams on his brakes to avoid striking a second yellow car in front of him, which had come to rest because of blocking ahead as shown in above Fig. After the brakes are applied, a constant friction force of 7.5 × 103 N acts on the car. Ignore air resistance.

Part (a)

Determine the least distance should the brakes be applied to avoid a collision with the other vehicle?

Solution:.

Mass of Car =2.0 × 103kg

Velocity of car=45m/s

Determine the least distance=s=d=?

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∴ Ffr = Âµk.N = ÂµK.mg

friction force = Normal force in opposite direction

Fn =− Ffr

−Î¼k .mg = m mg a = Î¼k .g

(g = 9.8 and Î¼ = 7.5×103 )

a = 7.5×103.(9.8)

a= -73.5×103

We know that

Newton 3rd law

v2 = u2 + 2as eq(1)

by using eq(1) value find out and putting value 0=(45)2 + 2(−73.5×103)s

2(−73.5×103)s = (45)2

Part (b)

If the distance between the vehicles is initially only at what speed would the collision occur?

What speed would the occur with velocity = 40.0 m?

Solution:

vi = 40m / s

Formula

v2= u2 + 2as

Putting values

V 2 = (40)2 + (−2013.9)

V 2 = 1600 − 2013.9

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Part (c)

Write your conclusive observations on the result obtained from this numerical. i.e. the importance of Physics in daily life.

Solution:

We conclude result that law of newton’s play key role in our lifestyles and whilst we appearance deep this scenario in line with physics then we can also see the importance in daily lifestyles. we can see the roads and motors also are designed in line with terms/guidelines of physics, for instance, the exits of the motorways are designed in calculated curve via way of means of the usage of the prescribed guidelines of centripetal force. The design of the motors additionally complies with the type. guidelines of physics which permit the car to journey quicker in opposition to the air.

Question 2:

Consider an automobile moving at v mph that skids d feet after its brakes lock. Calculate how far it would skid if it was moving at 2v and the brakes were locked.

Solution:

vi = Vmph

Vi = 0.44704 m/s

S= d feet means 1 feet convert into meters S=0.3048m

Vf2=vi2+2as (0)2=(0.44704)2+2(a)(0.3048)

=0.1998+2(0.3048) a

-0.1998=2(0.3048) a

a = −0.1998

2(0.3048)

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Velocity is 2v then we find the distance

Vf2-vi2=2as

0-(2)2=2(-0.3278) S

−4 = s

−2(0.3278)

2 = S

0.3278

s = 6.101

PLEASE NOTE:

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