# Question on work energy and power kinetic energy of an ....

1. Aug 18, 2016

### Mr.maniac

1. The problem statement, all variables and given/known data
If the speed of a vehicle increases by 2m/s then it's kinetic energy is doubled. Find out the original speed of the vehicle

2. Relevant equations
1/2(mv^2)

3. The attempt at a solution
First find out both KE(kinetic energy) taking u(initial vel.)
And v(final vel.) And mass as m
K(initial KE)= 1/2(mu^2)
K(final KE)=1/2(mv^2)
=1/2[m(u+2)^2]. {Given "speed of vehicle increases by 2m/s}

Also
2K=K
Then equating
2[1/2(mu^2)]=1/2[m(u+2)^2].
=>mu^2=1/2(mu^2+4m+4mu)
=>u^2=1/2(u^2+4+4u) {dividing m on both sides}
=>2u^2=u^2+4+4u
=>0= -u^2+4+4u.
We get
u=(√2+1)÷2
Sadly that ain't there in the options so. :~( {booo}

2. Aug 18, 2016

### SammyS

Staff Emeritus

3. Aug 18, 2016

### Mr.maniac

Will do

4. Aug 18, 2016

### Mr.maniac

OK redid it made a slight mistake but it's still no good

u=(-1+/-√2)÷2

5. Aug 18, 2016

### Mr.maniac

The +/- means plus or minus or this ± sorry didn't notice the sign before

6. Aug 18, 2016

### SammyS

Staff Emeritus
That's also wrong.

(The initial try looked better.)

7. Aug 18, 2016

### Staff: Mentor

(Edit: SammyS beat me to it!)

8. Aug 18, 2016

### Mr.maniac

OK got the mistake the ans now is 2(√2+1)
Which is correct I was taking the denominator wrong thank you very much for pointing that out

9. Aug 18, 2016

### Mr.maniac

Thank you all for helping