Question

A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE HELP

Answer 1

4978.09 kW

Step-by-step explanation:

Given, the mass of the car
`m=1430\textrm{ }kg`. The car speeds up from
`7.50\textrm{ }(m)/(s)\textrm{ to }11.0\textrm{ }(m)/(s)\textrm{ in }9.30\textrm{ s.}`.

Since friction is assumed zero, it is clear that all the work done by the forces responsible is going into increasing the speed of the car.

`\textrm{ Average Power = }\frac{\textrm{Total Work done}}{\textrm{Time taken}}`

`\textrm{Work done = Change in Kinetic Energy}`

`\textrm{Work done = }`Δ
`\textrm{ KE = }(1)/(2)mv_(f)^(2)-(1)/(2)mv_(i)^(2)`

`\textrm{Work done = }(1)/(2)(1430kg)(11.0(m)/(s))^(2)-(1)/(2)(1430kg)(7.50 (m)/(s))^(2)`
`\textrm{ = }46,296.25J`

`\textrm{Average Power = }(46296.25J)/(9.30s)=4978.09W=4.978kW`

∴ Power required =
`4.978kW`