Question

If a man speeds up by 1m/s, his kinetic energy increase by 44%. His original speed in m/s is

Answer 1

Answer:

Let the original speed of the man be “v₁” m/s.

So,  

K.E. = ½ * m * v₁² ……. (i)

After increasing the speed by 1 m/s:

Let the new speed of the man now be “v₂” = (v₁ + 1) m/s  

And  

New K.E. = K.E. + 44% of K.E = 144/100 K.E.

Therefore,  

New K.E. = ½ * m * v₂²  

⇒ (144/100) K.E.= ½ * m * (v₁+1)²

⇒ (144/100) * ½ * m * v₁² = ½ * m * (v₁+1)² ………. [from eq. (i)]

⇒ 144 v₁² = 100 (v₁+1)²

⇒ 144 v₁² = 100 (v₁² + 2v₁ + 1)

⇒ 144 v₁² = 100 v₁² + 200 v₁ + 100

⇒ 44v₁² -200 v₁ – 100 = 0

⇒ 11 v₁² – 55 v₁ – 25 = 0  

⇒ 11 v₁² – 55 v₁ + 5 v₁ -25 = 0  

⇒ 11v₁(v₁-5) + 5(v₁-5) = 0

⇒ v₁ = 5 m/s or v₁ = - 5/11 = - 0.4545  m/s

[Since value of v₁ cannot be negative here as the man is not moving in opposite direction in any of the cases]

∴ The original speed of the man is 5 m/s.

Answer 2

Answer:

Velocity of the man, v = 5 m/s.

Explanation:

In the question,

Let us say the speed of the man is  = v m/s

After speeding up,

New Velocity = (v + 1) m/s

Kinetic Energy (K.E.) is given by,

$KE=\frac{1}{2}mv^{2}$

New Kinetic Energy is 44% increased.

So,

$KE=KE +\frac{44}{100}(KE)=1.44KE$

So,

$1.44KE=\frac{1}{2}m(v+1)^{2}\\1.44\times \frac{1}{2}mv^{2}=\frac{1}{2}m(v^{2}+1+2v)\\1.44v^{2}=v^{2}+2v+1\\0.44v^{2}-2v-1=0\\11v^{2}-55v-25=0\\(v-5)(11v+5)=0\\v=5,\frac{-5}{11}$

Now, we know that the velocity of the person can not be negative.

So,

Velocity of the man is, v = 5 m/s.