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zhuklara [117]
3 years ago
12

What is the magnitude of the displacement of a car with an acceleration of 3.75 m/s2 as it increases its speed from 5.20m/s to 1

3.5m/s?
Physics
1 answer:
NeX [460]3 years ago
7 0

Answer:

20.7 m

Explanation:

using v^2=u^2+2as

s=(v^2-u^2)/2a

v=13.5

u=5.2

a=3.75

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We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
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While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

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⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

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⇒Fa*cos(20)+Fb*cos(50)=950

⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

⇒ Fb=\frac{950}{2.75}=345.5

⇒Fa=2.24*Fb

      =2.24*345.5

      =773.93

Therefore approximately, Fa=774 N and Fb=346 N

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