Ask your own question, for FREE!
Physics 13 Online
OpenStudy (unofficialllyy):

A car is driving at 20mph and has a kinetic energy of 1000J. If the car triples its speed, how much KE will it have?

OpenStudy (unofficialllyy):

@agent0smith

OpenStudy (unofficialllyy):

@IrishBoy123

OpenStudy (irishboy123):

what is the formula for KE?? d'ya know?

OpenStudy (unofficialllyy):

I believe it is \[\frac{ 1 }{ 2 }*mv^2\]

OpenStudy (irishboy123):

yes! so "initially" the car has KE: \(T_1 = \frac{1}{2} m v^2\) then it speeds up such that \(v \to 3v\) so it ends up with KE \(T_2 = \frac{1}{2}m (3 v)^2\) follow?

OpenStudy (unofficialllyy):

Yup!

OpenStudy (irishboy123):

So \(\Large T_2 = \frac{1}{2}m (3 v)^2 = 9 \times \frac{1}{2}m v^2 = 9 \times ?\)

OpenStudy (irishboy123):

put it another way we have concluded that \(\Large T_2 = ?? \times T_1\)

OpenStudy (agent0smith):

^

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!