Recent content by students2

Thanks a lot!

Yup, i just made all calculations. (I used meters per second instead of km's per hour). I am not going to write it down, but I calculated that accelerating from 27 kmph to 54 kmph will use more KE versus 0 to 27. So, if we equalize KE to work, than car must do more work going from 27 to 54, right?

Yes, exactly But now it has confirmed that velocity will be higher while accelerating from 27 to 54 versus 0 to 27, right? So, if I put that unknown velocity value in equation as v2, change in kinetic energy (E) from 27 to 54 is also going to be higher versus 0 to 27?

I mean, isn't velocity going to be higher when driving at 54 kmph? So, if the velocity doesn't changes (value stays the same) and E=½mv2 is the equation, then, change in kinetic energy also is the same?

Mass isn't given but velocity is going to be higher to reach 57 kmph from 27 kmph, right?

Nope, never really

I think it's those equations: 1) Fs = mdv / dt 2) v = ds / dt but I have no clue how to use them.

Homework Statement In which case car engine must do more work: To start driving from 0 kmph until car reaches 27 kmph or to reach 54 kmph from 27 kmph? Force of resistance in both cases are equal. I need to base answer of this problem on formulas. I apologize for any grammar mistakes (i'm not...