Question

The speed of a body moving with uniform acceleration is u this speed is doubled while covering a distance s. when it covers an additional distance s its speed would become?

Answer 1

Answer: The final speed of the body is $v_f  =  \sqrt 7 u.$

Given data states that the initial speed is u and a distance s is covered after which the speed is doubled that is it becomes 2u and covers s distance again. Therefore, from equation of motion,

                           $v^2 = u^2 + 2as\\ => a = v_i^2 – u^2 / 2s\\ => a = 2 u^2 – u^2 / 2s\\ => a = 3 u^2 / 2 s .$

Therefore, the final speed,

                           $v_f$ will be, $-v_f^2 -v_i^2 = 2 a s \\=> v_f^2 -4 u^2 = 2 \times (3 u^2/2 s) \times s \\=> v_f  =  \sqrt 7 u.$

Answer 2

Speed is  √7v  when it covers an additional distance S if  speed of a body 'v' moving with uniform acceleration is doubled in covering a distance S

Given:

• the speed of a body 'v' moving
• uniform acceleration
• Speed doubled in covering a distance S

To Find:

• Speed when it covers an additional distance S

Solution:

Step 1:

Calculate a

v² - u² = 2as

u = v

v = 2v  ( doubled)

s = S

=> (2v)² - v² = 2aS

=> 3v² = 2aS

=> a = 3v²/2S

Step 2:

Assume that final speed is x  then total distance covered = S + S= 2S

and solve for x  by substituting a =3v²/2S

x² - v²= 2a(2S)

=> x² - v²= 2(3v²/2S)(2S)  

=>  x² - v²=  6v²

=>  x² = 7v²

=> x = √7 v

Speed is  √7v  when it covers an additional distance S

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