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PHY101 ASSIGNMENT 1 SOLUTION 2021 | PHY101 ASSIGNMENT 1 FALL 2021 | PHY101 ASSIGNMENT 1 2021 | AN INTRODUCTION TO PHYSICS | VuTech

 

PHY101 ASSIGNMENT 1 SOLUTION 2021 | PHY101 ASSIGNMENT 1 FALL 2021 | PHY101 ASSIGNMENT 1 2021 | AN INTRODUCTION TO PHYSICS | VuTech

PHY101 ASSIGNMENT 1 SOLUTION 2021 | PHY101 ASSIGNMENT 1 FALL 2021 | PHY101 ASSIGNMENT 1 2021 | AN INTRODUCTION TO PHYSICS | VuTech

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Scenario:

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.

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

(b) If the distance between the vehicles is initially only 40.0 m, at what speed would the collision occur?

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

Question No. 1

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: 

Mass of Car = m = 2.0 x 103 kg

 

Velocity of Car = Vi = 45 m/s

 

Final Velocity = Vf = 0

 

Force of Friction = Fr = 7.5 x 103 kg

 

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

Least Distance = s = d = ?

 

As We know that

 

Ffr = mk ´ N = mk .mg

 

Friction Force = Normal Force in Opposite Direction

 

Fn = -Ffr

 

mg = - mk .mg

 

Acceleration = a = -mk .g

 

We know that

 

g = 9.8 m/s2   and   m =  7.5 x 103 kg

 

Acceleration = a = 7.5 x 103 x 9.8

 

a = -73.5 x 103

According to Newton’s 3rd Law,

 

V2 = u2 + 2as ------------------------(1)

 

Put all values in equation (1)

 

0 = (45)2 + 2 (-73.5 x 103)

 

S = 0.0137m Ans

 

(b) If the distance between the vehicles is initially only 40.0 m, at what speed would the collision occur?

If velocity is 40.0m, then
 
Vi = u = 40 m/s
 
According to Newton 3rd Law,
 
V2 = u2 + 2as
 
V2 = (40)2 + 2 (-73.5 x 103) (0.0137)
 
V2 = 1600 – 2013.9
 
V = -20.34 m/s Ans

 

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

Ans:

Physics, or the study of the elements, energy, and interactions between us helps us to understand the laws and laws that govern the universe. Not every student will grow up and learn physics at a deeper level, but everyone will use the basic concepts of physics to move around in daily life.


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Question No. 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.

Ans:

Vi  = Vmph

 

First we convert it into mph and Feet


Vi = 0.44704 m/s

 

S = d feet

 

(0)2 = (0.44704)2  + 2a(0.30480)

 

0 = 0.1998 + 2 (0.30480)a

 

a = -0.3278

 

If Velocity is 2V, then

 

= 2as

 

0 – (2)2 = 2 (-0.3278)s


S = 6.101m Ans

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